Vibrations and Waves MP205, Assignment 1 * 1. Consider a vector z defined by the equation z = z1 z2 , where z1 = a + ib, z2 = c + id. (a) Show that the length of z is the product of the lengths of z1 and z2 . (b) Show that the angle between z and the x axis is the sum of the angles made by z1 and z2 separately. (a) z = z1 z2 = (a + ib)(c + id) = (ac p |z1 | = a2 + b2 p |z2 | = c2 + d2 p p ) |z1 ||z2 | = a2 + b2 c2 + d2
bd) + i(ad + bc)
Now we find the length of z = z1 z2 . p p |z| = (ac bd)2 + (ad + bc)2 = a2 c2 + b2 d2 + a2 d2 + b2 c2 p p p = (a2 + b2 )(c2 + d2 ) = a2 + b2 c2 + d2 = |z1 ||z2 | (b) From the diagrams we see that:
tan ✓1 =
b a
,
tan ✓2 =
d c
We know that: a = |z1 | cos ✓1 b = |z1 | sin ✓1
c = |z1 | cos ✓2 d = |z1 | sin ✓2
Similarly, for z = z1 z2 we see that tan
=
ad + bc cos ✓1 sin ✓2 + sin ✓1 cos ✓2 sin(✓1 + ✓2 ) = = = tan(✓1 + ✓2 ) ac bd cos ✓1 cos ✓2 sin ✓1 sin ✓2 cos(✓1 + ✓2 )
2. Consider a vector z defined by the equation z = z1 /z2 , (z2 6= 0), where z1 = a + ib, z2 = c + id. (a) Show that the length of z is the quotient of the lengths of z1 and z2 . [8] (b) Show that the angle between z and the x axis is the di↵erence of the angles made by z1 and z2 separately. [7] (a) This follows trivially from the fact that: z1 |z1 | = z2 |z2 |
|z| = Alternatively we can do it out:
z1 a + ib (a + ib)(c id) (ac + bd) + i(bc = = = 2 2 z c + id c +d c2 + d 2 s2 (ac + bd)2 + (bc ad)2 |z| = (c2 + d2 )2 p a2 c2 + 2acbd + b2 d2 + b2 c2 2abcd + a2 d2 = c2 + d 2 p a2 + b2 (c2 + d2 ) = c2 +pd2 p a2 + b 2 c 2 + d 2 = 2 2 p c +d a2 + b 2 =p c2 + d 2 |z1 | = |z2 | z=
(b) Drawing out z1 and z2
We know that: a = |z1 | cos ✓1 b = |z1 | sin ✓1
c = |z1 | cos ✓2 d = |z1 | sin ✓2 Im
z φ
bc-ad c2 + d 2
ac+bd Re c2 + d 2
ad)
From the diagram above and using the values of a, b, c, d we get: ✓ ◆✓ 2 ◆ bc ad c + d2 bc ad sin ✓1 cos ✓2 cos ✓1 sin ✓2 tan ' = = = 2 2 c +d ac + bd ac + bd cos ✓1 cos ✓2 + sin ✓1 sin ✓2 sin(✓1 ✓2 ) = = tan(✓1 ✓2 ) cos(✓1 ✓2 ) ) ' = ✓1 ✓2 3. Show that the multiplication of any complex number z by ei✓ is describable, in geometric terms, as a positive rotation through the angle ✓ of the vector by which z is represented, without any alteration of its length. [5] Consider the complex number z, that makes an angle with the x axis such that z = a + ib = |z|ei . We multiply z by ei✓ , that is, zei✓ = |z|ei ei✓ = |z|ei( +✓) . We see that there is no alteration of its length as before and after multiplication its length is |z|. It has however undergone a positive rotation of ✓ as we can see from the diagrams below. Im
Im
|z|
|z| φ+θ
φ
Re
Re
* 4. Would you be willing to pay 20 cents for any object valued by a mathematician at ii cents? Evaluate Euler’s relation ei✓ = cos ✓ + i sin ✓, at ✓ = ⇡/2. This gives ei⇡/2 = cos ⇡/2 + i sin ⇡/2 = i. However since cos(✓) = cos(✓ + 2n⇡) and sin(✓) = sin(✓ + 2n⇡), for n = 0, 1, 2, ..., we know that ei✓ = ei(✓+2n⇡) . So we can write i = ei(⇡/2+2n⇡) We want to find the value of ii , hence ii = (ei(⇡/2+2n⇡) )i = ei
2 (⇡/2+2n⇡)
=e
(⇡/2+2n⇡)
Looking at this for the first 3 values of n: • n = 0: e
(⇡/2+2n⇡)
=e
⇡/2
(⇡/2+2n⇡)
=e
(⇡/2+2⇡)
• n = 2: e
(⇡/2+2n⇡)
=e
(⇡/2+4⇡)
• n = 1: e
⇡ 21c
⇡ 0.04c
⇡ 0.00007c
So the answer here is maybe, it depends on your value of n! 5. (a) If z = Aei✓ , deduce that dz = izd✓, and explain the meaning of this relation in a vector diagram. [8] p (b) Find p 2 the magnitudes and directions of the vectors (2 + i 3) and (2 i 3) . [10]
(a) z = Aei✓
dz = iAei✓ d✓
,
dz = iAei✓ d✓ = izd✓
)
Multiplication by i shifts phase by ⇡/2 as depicted below. Im Im
z
dz
θ
θ + π/2
Re
(b) We find the magnitudes (lengths) firstly. q p p |z1 | = 22 + 32 = 7 p p z2 = (2 i 3)2 = 1 i4 3 )
Re
q p |z2 | = 12 + ( 4 3)2 = 7
By drawing both complex numbers, we can easily calculate their directions. Im Im z1 Re θ1
Re
2
1
z2
p For z1 , we have ✓1 = tan 1 (y/x) = tan 1 ( 3/2) = 40.89 .p For z2 , we have ✓2 = 360 tan 1 (y/x) = 360 tan 1 (4 3/1) = 278.21 . 6. Given Euler’s relation ei✓ = cos ✓ + i sin ✓, find (a) An expression for e
i✓
. [5]
(b) The exponential representation of cos ✓. [5] (c) The exponential representation of sin ✓. [5] (a) e
i✓
= cos( ✓) + i sin( ✓) = cos(✓) i sin(✓)
(b) We have that ei✓ = cos ✓ + i sin ✓ and e expressions together: ei✓ + e
i✓
= cos ✓
i✓
= 2 cos ✓ 1 ) cos ✓ = (ei✓ + e 2
(c) Similarly, using ei✓ = cos ✓ + i sin ✓ and e ei✓
e
i sin ✓. Adding both
i✓
i✓
)
= cos ✓
i✓
= 2i sin ✓ 1 ) sin ✓ = (ei✓ e 2i
i✓
)
i sin ✓, then:
* 7. Justify the formulas cos ✓ = (ei✓ + e i✓ )/2 and sin ✓ = (ei✓ appropriate series. The series expansion for ex is given by: e =
n=0
i✓
i✓
)/2i, using the
n
we get:
ei✓ = 1 + e
i✓
1 X xn
x
By expanding ei✓ and then e
e
=1
✓2 i✓3 ✓4 + + ... 2! 3! 4! ✓2 i✓3 ✓4 + + + ... 2! 3! 4!
i✓ 1! i✓ 1!
Therefore the series expansion for cos ✓ is cos ✓ =
ei✓ + e 2
i✓
=1
✓2 ✓4 + + ... 2! 4!
which can be verified by Taylor expanding cos ✓. The Taylor expansion a function f (x) is given by: f (x) = f (0) +
xf 0 (0) x2 f 00 (0) xn f n (0) + + ... + 1! 2! n!
Taylor expanding cos ✓ we get: ✓2 ( cos(0)) ✓3 (sin(0)) ✓4 ( cos(0)) cos ✓ = cos(0) + ✓( sin(0)) + + + + ... 2 3 4 ✓2 ✓4 =1 + + ... 2 4 Which is what got above. Similarly the series expansion for sin ✓ is sin ✓ =
ei✓
e 2i
i✓
=✓
✓3 ✓5 + + ... 3! 5!
which can be verified by Taylor expanding sin ✓. Taylor expanding cos ✓ we get: sin ✓ = sin(0) + ✓(cos(0)) + =✓
✓2 ( sin(0)) ✓3 ( cos(0)) ✓4 (sin(0)) ✓5 (cos(0)) + + + ... 2 3 4 4
✓3 ✓5 + + ... 3 5
Which is what got above. 8. To take successive derivatives of ei✓ with respect to ✓, one merely multiplies by i: d (Aei✓ ) = iAei✓ d✓ Show that this prescription works if the sinusoidal representation ei✓ = cos ✓ + i sin ✓ is used. [5]
Let f (✓) = cos ✓ + i sin ✓. Want to show that: d f (✓) = if (✓). d✓ Evaluating the LHS gives: d (cos ✓ + i sin ✓) = d✓
sin ✓ + i cos ✓
The RHS can be rewritten as: i(cos ✓ + i sin ✓) = i cos ✓ + i2 sin ✓ = i cos ✓
sin ✓
Clearly on the LHS we have the same expression as on the RHS, hence d f (✓) = if (✓). d✓ 9. Using the exponential representations for sin ✓ and cos ✓, verify the following trigonometric identities: (a) sin2 ✓ + cos2 ✓ = 1 [6] sin2 ✓ = cos 2✓ [6]
(b) cos2 ✓
(c) 2 sin ✓ cos ✓ = sin 2✓ [6] We know that cos ✓ = (ei✓ + e i✓ )/2 and sin ✓ = (ei✓ e (e2i✓ + 2 + e 2i✓ )/4 and sin2 ✓ = (e2i✓ 2 + e 2i✓ )/4.
i✓
)/2i, therefore cos2 ✓ =
(a) e2i✓
sin2 ✓ + cos2 ✓ =
2+e 4
2i✓
2i✓
e2i✓
+
e2i✓ + 2 + e 4
2i✓
=
1 1 + =1 2 2
(b) cos2 ✓
sin2 ✓ =
e2i✓ + 2 + e 4
+
2i✓
2+e 4
=
e2i✓ + e 2
2i✓
= cos 2✓
(c) 2 sin ✓ cos ✓ = 2
✓
ei✓
e 2i
i✓
◆✓
ei✓ + e 2
i✓
◆
* 10. Verify that the di↵erential equation d2 y/dx2 =
=
e2i✓
e 2i
2i✓
= sin 2✓
k 2 y has as its solution
y = A cos(kx) + B sin(kx), where A and B are arbitrary constants. Show also that this solution can be written in the form y = C cos(kx + ↵) = CRe[ei(kx+↵) ] = Re[(Ce(i↵) )eikx ] and express C and ↵ as functions of A and B.
11. A mass on the end of a spring oscillates with an amplitude of 5cm at a frequency of 1Hz (cycles per second). At t = 0 the mass is at its equilibrium position (x = 0). (a) Find the possible equations describing the position of the mass as a function of time, the form x = A cos(!t + ↵), giving the numerical values of A, !, and ↵. [6] (b) What are the values of x, dx/dt, and d2 x/dt2 at t = 8/3 seconds? [6] (a) At time t = 0 we have x = 0. Filling into the equation of motion gives 0 = cos ↵ ) ↵ = ±⇡/2. Frequency is 1Hz ) Period T = 1/F = 1 second. Angular frequency ! = 2⇡/T = 2⇡s 1 . Amplitude A is given in the question as 5cm. This gives two possible equations describing the position of the mass as a function of time: ⇣ ⇡⌘ x(t) = 5 cos 2⇡t + 2 and ⇣ ⇡⌘ x(t) = 5 cos 2⇡t 2 (We’d need an initial velocity to work out a specific case) (b) Looking at x(t) = 5 cos 2⇡t + ⇡2 : p ✓ ◆ ✓ ◆ ⇣⇡ ⌘ 8 ⇡ 11⇡ 3 x(8/3) = 5 cos 2⇡ + = 5 cos = 5 cos =5 cm 3 2 6 6 2 dx/dt = !A sin(!t + ↵) ✓ ◆ ✓ ◆ dx 8 ⇡ 11⇡ (8/3) = 2⇡(5) sin 2⇡ + = 10⇡ sin = 5⇡ cms 1 dt 3 2 6 d2 x/dt2 = ! 2 A cos(!t + ↵) ✓ ◆ ⇣⇡ ⌘ p d2 x 8 ⇡ 2 2 (8/3) = (2⇡) (5) cos 2⇡ + = 20⇡ cos = 10⇡ 2 3 cms 2 dt 3 2 6
2
12. A point moves in a circle at a constant speed of 50cm/sec. The period of one complete journey around the circle is 6 seconds. At t = 0 the line to the point from the center of the circle makes an angle of 30 with the x axis. (a) Obtain the equation of the x coordinate of the point as a function of time, in the form x = A cos(!t + ↵), giving the numerical values of A, !, and ↵. [6] (b) Find the values of x, dx/dt, and d2 x/dt2 at t = 2 seconds. [6] (a) Since at t = 0 the line to the point from the center of the circle makes an angle of 30 with the x axis, this means that the initial phase ↵ = ⇡/6. Now: distance 2⇡r 2⇡A ) 50 = = time T 6 2⇡ ⇡ 1 != = s T 3
velocity =
)A=
150 cm ⇡
(b) x = A cos(!t + ↵) = (150/⇡) cos(⇡t/3 + ⇡/6) p ✓ ◆ ⇣⇡ ⌘ 150 5⇡ 150 75 3 x(2) = cos = cos = cm ⇡ 6 ⇡ 6 ⇡ dx/dt = !A sin(!t + ↵) ✓ ◆ ⇣⇡ ⌘ dx ⇡ 150 5⇡ ⇡ 150 (2) = . sin = . sin = 25 cms dt 3 ⇡ 6 3 ⇡ 6 d2 x/dt2 = ! 2 A cos(!t + ↵) ✓ ◆ ⇣ ⇡ ⌘ 25⇡ d2 x ⇡ 2 150 5⇡ ⇡ 2 150 (2) = . cos = . cos = p cms dt2 9 ⇡ 6 9 ⇡ 6 3
1
2
Vibrations and Waves MP205, Assignment 2 Solutions 1. Express the following in the form Re[Aei(!t+↵) ] (a) z = sin !t + cos !t. (b) z = cos(!t
⇡/3)
cos !t.
(c) z = 2 sin !t + 3 cos !t. (d) z = sin !t
2 cos(!t
⇡/4) + cos !t.
First recall that Aei(!t+↵) can be written in two ways: Aei(!t+↵) = A{cos(!t) + i sin(!t)}{cos(↵) + i sin(↵)} = A{cos(!t) cos(↵) sin(!t) sin(↵)} + Ai{cos(!t) sin(↵) + sin(!t) cos(↵)} or Aei(!t+↵) = A{cos(!t + ↵) + i sin(!t + ↵)} From this we can write Re[Aei(!t+↵) ] as Re[Aei(!t+↵) ] = A cos(!t) cos(↵)
A sin(!t) sin(↵)
or Re[Aei(!t+↵) ] = A cos(!t + ↵) For these solutions I’ll do the first format. (a) z = sin !t + cos !t sin(!t) + cos(!t) = A cos(!t) cos(↵) 1=
A sin(↵) 1 A= sin(↵) A is positive so we know sin(↵) < 0
A sin(!t) sin(↵) 1 = A cos(↵) 1 A= cos(↵) cos(↵) > 0
1 1 = sin(↵) cos(↵) sin(↵) = 1 cos(↵)
↵=
7⇡ 4
p 1 1 = = 2 cos(↵) cos 7⇡ 4 p 7⇡ ) z = Re[ 2ei(!t+ 4 ) ] A=
(b) z = cos(!t
⇡/3)
cos !t = cos(!t
⇡/3) + cos(!t + ⇡).
z = cos(!t
⇡3) cos(!t) ⇣⇡ ⌘ ⇣⇡ ⌘ cos(!t ⇡3) = cos(!t) cos + sin(!t) sin 3 3 p 1 3 = cos(!t) + sin(!t) 2 2 p 1 3 ) z = cos(!t) + sin(!t) cos(!t) 2 2p 1 3 = cos(!t) + sin(!t) 2 2 p
3 = 2
2A =
1 = A cos(↵) 2 1 2A = cos(↵)
A sin(↵) p 3 sin(↵)
A is positive so we know sin(↵) < 0
cos(↵) < 0 p
3 1 = sin(↵) cos(↵) p sin(↵) = 3 cos(↵)
↵=
4⇡ 3
2A =
1 1 = =2 cos(↵) cos 4⇡ 3
)A=1
) z = Re[ei(!t+ 3 ) ]
(c) z = 2 sin !t + 3 cos !t = 2 cos(!t
4⇡
⇡/2) + 3 cos(!t).
z = 2 sin(!t) + 3 cos(!t) 2=
A sin(↵) 2 A= sin(↵) A is positive so we know sin(↵) < 0
3 = A cos(↵) 3 A= cos(↵) cos(↵) > 0 2 3 = sin(↵) cos(↵) sin(↵) 2 = cos(↵) 3
↵ = 5.7rad 3 A= = cos(↵) cos 2⇡ p i(!t+5.7) ) z = Re[ 13e ] (d) z = sin !t
2 cos(!t
3 tan
1
=
2 3
p
13
⇡/4) + cos !t
⇣ ⇡⌘ z = sin(!t) 2 cos !t + cos(!t) 4 ⇣ ⇣⇡ ⌘ ⇣⇡ ⌘ ⇡⌘ cos !t = cos(!t) cos + sin(!t) sin 4 4 4 1 1 = p cos(!t) + p sin(!t) 2 2 p p ) z = sin(!t) 2 cos(!t) 2 sin(!t) + cos(!t) p p = (1 2) cos(!t) + (1 2) sin(!t) (1
p
2) =
p
A sin(↵)
(1
(1 2) sin(↵) A is positive so we know sin(↵) > 0 A=
p
2) = A cos(↵) p (1 2) A= cos(↵) cos(↵) < 0
p p (1 2) (1 2) = sin(↵) cos(↵) sin(↵) = 1 cos(↵)
3⇡ 4 p p (1 2) (1 2) A= = =2 3⇡ cos(↵) cos 4 p i !t+ 3⇡ 4 )] ) z = Re[(2 2)e ( ↵=
p
2
* 2. A particle is simultaneously subjected to three simple harmonic motions, all of the same frequency and in the x direction. If the amplitudes are 0.25, 0.20, and 0.15 mm, respectively, and the phase di↵erence between the first and second is 45 , and between the second and third is 30 , find the amplitude of the resultant displacement and its phase relative to the first (0.25 mm amplitude) component. So again we want to get this in the form: Re[Aei(!t+↵) ] = A cos(!t) cos(↵)
A sin(!t) sin(↵)
First we need to set up our system: x1 = 0.25 cos(!t) ⇣ ⇡⌘ x2 = 0.20 cos !t + 4⇣ ⌘ h ⇣ ⇡ ⌘i ⇡ = 0.20 cos(!t) cos sin(!t) sin 4 4 1 1 = 0.20 p cos(!t) p sin(!t) 2 2 ⇡ 0.14 cos(!t) 0.14 sin(!t) ✓ ◆ ⇣ ⇡ ⇡⌘ 5⇡ x3 = 0.15 cos !t + + = 0.15 cos !t + 4 6 12 ✓ ◆ ✓ ◆ 5⇡ 5⇡ = 0.15 cos(!t) cos sin(!t) sin 12 12 ⇡ 0.15 [0.26 cos(!t) 0.97 sin(!t)] = 0.039 cos(!t) 0.136 sin(!t) Our particle is simultaneously to these three simple harmonic motions: x = x1 + x2 + x3 = 0.25 cos(!t) + 0.14 cos(!t) = 0.43 cos(!t) 0.28 sin(!t)
0.14 sin(!t) + 0.039 cos(!t)
0.136 sin(!t)
0.28 = A sin(↵) 0.28 A= sin(↵) A is positive so we know sin(↵) > 0
0.43 = A cos(↵) 0.43 A= cos(↵) cos(↵) > 0
0.28 0.43 = sin(↵) cos(↵) sin(↵) 0.28 = = 0.65 cos(↵) 0.43
↵ = 0.58rad 0.43 0.43 A= = = 51mm cos(↵) cos(0.58) ) z = Re[51ei(!t+0.58) ] 3. Two vibrations along the same line are described by the equations y1 = A cos(10⇡t) y2 = A cos(12⇡t) Find the beat period, and draw a careful sketch of the resultant disturbance over one beat period.
beat period T =
2⇡ |!1
!2 |
=
2⇡ |10⇡
12⇡|
= 1s
4. Find the frequency of the combined motion of each of the following: p * (a) sin(2⇡t 2) + cos(2⇡t). (b) sin(12⇡t) + cos(13⇡t (c) sin(3t)
⇡/4).
cos(⇡t).
Here we use the fact that when two SHM’s are quite close in frequency, they have a frequency equal to the average of the combining frequencies (but with an amplitude that varies periodically with time). Also recall the following formulae: 1 T 2⇡ T = ! 1 ! )f = = T 2⇡ f=
(a) sin(2⇡t
p
2) + cos(2⇡t). x = x1 + x2 = sin(2⇡t x1 = sin(2⇡t ⇣ = cos 2⇡t
p p
2) + cos(2⇡t)
2) p 2
) !1 = 2⇡ !1 2⇡ ) f1 = = =1 2⇡ 2⇡ x2 = cos(2⇡t) ) !2 = 2⇡ !2 2⇡ ) f2 = = =1 2⇡ 2⇡ The average of these frequencies is given by: f= (b) sin(12⇡t) + cos(13⇡t
⇡⌘ 2
f1 + f2 1+1 = =1 2 2
⇡/4).
x = x1 + x2 = sin(12⇡t) + cos(13⇡t x1 = sin(12⇡t) ⇣ ⇡⌘ = cos 12⇡t 2 ) !1 = 12⇡ !1 12⇡ ) f1 = = =6 2⇡ 2⇡ x2 = cos(13⇡t ⇡/4) ) !2 = 13⇡ !2 13⇡ ) f2 = = = 6.5 2⇡ 2⇡ The average of these frequencies is given by: f=
f1 + f2 6 + 6.5 = = 6.25 2 2
⇡/4)
(c) sin(3t)
cos(⇡t). x = x1 + x2 = sin(3t) x1 = sin(3t) ⇣ ⇡⌘ = cos 3t 2 ) !1 = 3 3 ) f1 = 2⇡ x2 = cos(⇡t) = cos(⇡t + ⇡) ) !2 = ⇡ !2 ⇡ 1 ) f2 = = = 2⇡ 2⇡ 2
cos(⇡t)
The average of these frequencies is given by: f=
f1 + f2 = 2
3 2⇡
+ 2
1 2
⇡ 4.9
* 5. Two vibrations at right angles to one another are described by the equations x = 10 cos(5⇡t) y = 10 cos(10⇡t + ⇡/3) Construct the Lissajous figure of the combined motion. We have x = 10 cos(5⇡t), y = 10 cos(10⇡t + ⇡/3).
6. Construct the Lissajous figures for the following motions: (a) x = cos(2!t), y = sin(2!t). (b) x = cos(2!t), y = cos(2!t
⇡/4).
(c) x = cos(2!t), y = cos(!t). (a) We have x = cos(2!t), y = sin(2!t) = cos(2!t
(b) We have x = cos(2!t), y = cos(2!t
(c) We have x = cos(2!t), y = cos(!t).
⇡/4).
⇡/2).
Vibrations and Waves MP205, Assignment 3 Solutions 1. An item of mass 1 g is hung from a spring and set in oscillatory motion. At t = 0 the displacement is 43.785 cm and the acceleration is -1.7514cm/sec2 . What is the spring constant? Given m = 1 g = 1 ⇥ 10 3 kg and at t = 0, x = 43.785 cm = 0.43785 m and a = 1.7514cms 2 = 0.017514 ms 2 . Undergoes oscillatory motion so it must satisfy an equation of the form a = ! 2 x. !2x k = x m am k= x ( 0.017514)(1 ⇥ 10 3 ) = 0.43785 5 = 4 ⇥ 10 Nm 1 a=
* 2. A mass m hangs from a uniform spring of spring constant k. (a) What is the period of oscillations in the system? (b) What would it be if the mass m were hung so that: (see Figure below) (1) It was attached to two identical springs hanging side by side? (2) It was attached to the lower of two identical springs connected end to end?
k
k k
k (1)
(a) F =
kx, T =
2⇡ !
and ! =
(2)
q
k . m
Then
T = 2⇡
r
m k
(b)(1) To move mass an equivalent displacement x as in part (a) we need twice the force, since the restoring force is now twice as big i.e. k 0 = 2k. This means that r
r 0 k 2k !0 = = m rm 2⇡ m T T 0 = 0 = 2⇡ =p ! 2k 2
(b)(2) k is inversely proportional to the length of the string. Let l be the length of the original spring, then: k/
1 l
k=
a l
or equivalently for some constant a The length of our spring here is 2l k 00 = This means that r
k 00
a k = 2l 2
s
k 2
r
k m m 2m r p 2⇡ 2m T 00 = 00 = 2⇡ =T 2 ! k ! 00 =
=
=
3. A platform is executing simple harmonic motion in a vertical direction with an amplitude of 5cm and a frequency of 10 vibrations per second. A ⇡ block is placed on the platform at the lowest point of its path (a) At what point will the block leave the platform? [Hint: think of the forces on the block ] (b) How far will the block rise above the highest point reached by the platform? (a) We have the following information: A = 5cm 10 f= ⇡ ⇡ 2⇡ T = = 10 ! ! = 20 If we look at the forces on the block:
Relative to the block, the net force is 0 for as long as the block is on the platform. Analysing the above diagram, kx is the force exerted by the platform on the block (due to the simple harmonic motion), W is the weight of the block, and N is the force exerted by the block on the platform (the normal force-as kx > W . If kx < W then the system would stop oscillating when the block is place on the platform). This force N is the force keeping the block on the platform, it is NOT constant, and changes with respect to x. We want to find the point when N = 0 - this is the point the block leaves the platform. kx = W N kx = W looking at the point whereN = 0 kx = mg m x= g k m = g m! 2 g = 2 ! 9.8 = 400 = 0.0245m = 2.5cm (b) We need to find the velocity of the block as it leaves: x = A cos(!t) = 5 cos(20t) 2.5 = 5 cos(20t) 0.5 = cos(20t) ⇡ ) 20t = 3 dx v= = 5(20) sin(20t) = dt Looking at v when x leaves the platform: ⇣⇡ ⌘ v = 100 sin 3 = ±86.6cms 1
100 sin(20t)
Note: The ± comes from the fact we can take 20t = ⇡3 or 4⇡ - it just indicates 3 that in a SHM system we’ll have the same speed at a point at 2 di↵erent times, just in 2 di↵erent directions depending on the direction of the motion. Using the formula that relates vi , vf , a and s, where: • • • •
vi is the initial velocity vf is the final velocity a is the acceleration s is the displacement vf2 = vi2 + 2as
We know our initial velocity is vi = 86.6cms 1 (as our motion is up we take the plus direction), once the block leaves the platform its acceleration will just be gravity a = 9.8, and the velocity when the block hits its highest point is just vf = 0. vf2 = vi2 + 2as 0 = (86.6)2 + 2( 9.8)s 0.76 s= = 0.038m = 3.8cm 19.8 Max height of the platform is just the amplitude A = 5cm. The height the block reaches is its distance travelled on the platform 2.5cm and its distance travelled once it left 3.8cm. This gives the height above the platform reached as: h = (3.8 + 2.5)
5 = 1.3cm
* 4. A uniform rod of length L is nailed to a post so that two thirds of its length is below the nail. What is the period of small oscillations of the rod?
If we look at the period in terms of the center of mass, where the center of mass C is a distance h from the point of suspension, and k is the radius of gyration of the body. In this case, the period T is given by: ✓ 2 ◆1 h + k2 2 T = 2⇡ gh Looking at our system:
Our center of mass C will be at the center of the rod, so our h is given by h=
2L 3
L L = 2 6
To get k, we recall that I = mk 2 , and that the moment of inertia of a rod (through 1 its center of mass) is given by I = 12 mL2 . I = mk 2 I ) k2 = m L2 = 12 This gives us the period: L2 36
T = 2⇡
L2 12
gL 6 1 6
L
= 2⇡ = 2⇡
+
s
+ g
! 12
! 12
1 2
2L 3g
So we can think of the system as a simple pendulum of length simple pendulum is given by: s l T = 2⇡ g
2L . 3
The period of a
where l is the length of the pendulum. l=
2L 3s
) T = 2⇡
2L 3g
5. A circular hoop of diameter d hangs on a nail. What is the period of its oscillations at small amplitude? If we look at the period in terms of the center of mass, where the center of mass C is a distance h from the point of suspension, and k is the radius of gyration of the body. In this case, the period T is given by: T = 2⇡ Looking at our system:
✓
h2 + k 2 gh
◆ 12
Our center of mass will be at the center of the circle, so our h is just the radius of the circle, and as we have a circle our k is also just the radius. This gives us the period: ✓
r2 + r2 T = 2⇡ gr ✓ ◆ 12 2r = 2⇡ g r 2r = 2⇡ g
◆ 12
But 2r = d ) T = 2⇡
s
d g
We can think of the system as a simple pendulum of length d. The period of a simple pendulum is given by: s l T = 2⇡ g where l is the length of the pendulum. l=d ) T = 2⇡
s
d g
* 6. (a) An object of mass 0.5 kg is hung from the end of a steel wire of length 2 m and of diameter 0.5 mm. (Young’s modulus = 2 ⇥ 1011 N/m2 ). What is the extension of the wire? (b) The object is lifted through a distance h (thus allowing the wire to become slack) and is then dropped so that the wire receives a sudden jerk. The ultimate strength of steel is 1.1⇥109 N/m2 . What is the largest possible value of h if the wire is not to break?
(a) Y =
stress = strain
P A l l0
P = mg
(0.5)(9.8) = 4.9N ✓ ◆2 0.5 ⇥ 10 3 2 A = ⇡r = ⇡ = 1.96 ⇥ 10 7 m2 2 l0 = 2m
Y = 2 ⇥ 1011 Nm ) 2 ⇥ 1011 =
4.9 1.96⇥10 l 2
2
7
5 ⇥ 107 l 5 ⇥ 107 l= = 2.5 ⇥ 10 4 m = 0.25mm 2 ⇥ 1011
2 ⇥ 1011 = )
(b) Ultimate strength = max value of stress. We know that: Y =
stress strain
Y is constant, so if we have maximum stress, this corresponds to the maximum strain. To get the maximum extension l we need to get the maximum strain, as l / strain. max stress max strain 1.1 ⇥ 109 2 ⇥ 1011 = max strain 1.1 ⇥ 109 ) max strain = = 5.5 ⇥ 10 2 ⇥ 1011 max l = 5.5 ⇥ 10 3 l max l = 5.5 ⇥ 10 3 2 max l = 0.011m Y =
3
So the maximum change in length before the wire will break is 0.011m. To find the maximum height we can lift the object, we look at the energy of the system when lifted to a height h, and the energy of the system at the lowest point l0 . • At the highest point (h) the total energy is just the potential energy: mgh. • At the lowest point ( l) the total energy is just kinetic energy: 12 kx2 = 1 2 kl 2 0 Using conservation of energy: mgh = 12 k l2
k is just given by k =
AY l0
=
(1.96⇥10
7 )(2⇥1011 )
2
= 19600.
k l2 2mg (19600)(0.011)2 = 2(0.5)(9.8) = 0.24m
h=
7. A solid steel ball is to be hung at the bottom end of a steel wire of length 2m and radius 1mm. The ultimate strength (max stress) of steel is 1.1 ⇥ 109 N/m2 . What are: (a) the mass of the biggest ball the wire can bear (b) the radius of the biggest ball the wire can bear [Hint: use the fact that the density of steel is given by ⇢s = 7900kgm 3 ] (a) We know the ultimate strength of steel is 1.1 ⇥ 109 N/m2 , and r = 1mm = 0.001m: P A mg = 2 ⇡r m(9.8) = ⇡(0.001)2 3455.75 = 9.8m m = 352.6kg 1.1 ⇥ 109 =
(b) We know the density of steel is ⇢s = 7900kgm 3 , and m = V ⇢: m=V⇢ 4 = ⇡r3 ⇢ 3 3m r3 = 4⇡⇢ 3(352.6) = 4(3.14)(7900) = 0.0107 r = 0.22m 8. A metal rod, 0.5 m long, has a rectangular cross section of area 2 mm2 . With the rod vertical and a mass of 60kg hung from the bottom, there is an extension of 0.25 mm. What is the Young’s modulus ( N/m2 ) for the material of the rod? Given m = 60 kg, h = 0.25 mm = 0.25 ⇥ 10 2 ⇥ 10 6 m2 . We also know that mg =
3
m, `0 = 0.5 m and A = 2 mm2 =
AY h mg`0 (60)(9.8)(0.5) )Y = = = 5.88 ⇥ 1011 Nm `0 Ah (2 ⇥ 10 6 )(0.25 ⇥ 10 3 )
2
Vibrations and Waves MP205, Assignment 4 Solutions 1. (a) Show that the frequency of vibration under adiabatic conditions of a column of gas confined to a cylindrical tube, closed at one end, with a well-fitting but freely moving piston of mass m is given by: r A p != lm . (b) A steel ball of diameter 2cm oscillates vertically in a 12-liter glass tube containing air at atmospheric pressure (as pictured below). Verify that the period of oscillation should be about 1 sec. (Assume adiabatic pressure change with = 1.4. Density of steel = 7600kgm 3 .)
(a) Under adiabatic conditions we have :pV = constant pV = const ln(pV ) = const ln(p) + ln(V ) = const ln(p) + ln(V ) = const Di↵erentiating this w.r.t. V gives: 1 dp + =0 p dV V dp p + =0 dV V dV p dp = V Now we know we can express : • force F as F = A p (= Adp) • volume V as V = Al • the change in volume V as V = Ay (= dV )
F = Adp dV p = A V A2 y p = Al A p = y l A p ma = y l A p a= y ml We know a =
!2y A p ml r A p != ml
) !2 =
(b) 4 4 mass of ball;m = V ⇢ = ⇡r3 ⇢ = ⇡ 3 3 2 2 volume of tube;V = lA = l⇡r ✓ ◆2 0.02 0.012 = l⇡ 2 ) l ⇡ 38.2m
✓
0.02 2
◆3
(7600) ⇡ 0.032kg
Using this: !=
r
A p s ml
0.02 2 2
(1.4)(101325) p = 36.5 = 6.04sec (38.2)(0.032) 2⇡ 2⇡ )T = = = 1.04 ⇡ 1sec ! 6.04 =
⇡
1
* 2. The motion of a linear oscillator may be represented by means of a graph in which x is shown as abscissa and dx/dt as ordinate. The history of the oscillator is then a curve. (a) Show that for an undamped oscillator this curve is an ellipse. (b) Show (at least qualitatively) that if a damping term is introduced one gets a curve spiraling into the origin. (a) For an undamped oscillator we have x = A cos(!t + ↵). x = A cos(!t + ↵) dx = !A sin(!t + ↵) dt
We want to show that x = A cos(!t + ↵) and y = dx = !A sin(!t + ↵) satisfy dt an ellipse equation. 2 2 Ellipse equation is given by: xa + yb = 1 In our case x = A cos(!t + ↵) and y = dx = !A sin(!t + ↵) dt x2 x2 A2 y2 y2 ! 2 A2
= A2 cos2 (!t + ↵) = cos2 (!t + ↵) = ! 2 A2 sin2 (!t + ↵) = sin2 (!t + ↵)
Using: sin2 (x) + cos2 (x) = 1 We can write:
as required.
x2 y2 + =1 A2 ! 2 A2 ⇣ x ⌘2 ⇣ y ⌘2 + =1 A !A ) the curve is an ellipse
(b) If we think of our ellipse in (a) in terms of a pendulum:
• If we release the pendulum from the point 1, it’s velocity is 0 and it has maximum (positive) displacement • When the pendulum goes through 2 it has maximum (negative, as it is going in the negative direction) velocity and its displacement is 0 • When the pendulum reaches 3 it has maximum (negative) displacement, and 0 velocity • Finally, when the pendulum returns through 4 it will have maximum (positive) velocity, and again 0 displacement • We can see this corresponds to the points on the ellipse to the right
• If we release the pendulum from the point 1, it’s velocity is 0 and it has maximum (positive) displacement, as before. • When the pendulum goes through 2 it has its maximum (negative, as it is going in the negative direction) velocity for that swing (although this is less than the maximum velocity before due to damping) and its displacement is 0 • When the pendulum reaches 3 it has its maximum (negative) displacement for this swing, but due to damping this displacement is not as great as the original maximum displacement, and 0 velocity • When the pendulum returns through 4 it will have its maximum (positive) velocity for this swing, which is less than the velocity when it went through it on ”2 , and again 0 displacement • Continuing in this vein we can see the graph that corresponds to this is a curve spiralling into the origin. 3. Verify that x = Ae
↵t
cos !t is a possible solution of the equation d2 x dx + + !02 x = 0, 2 dt dt
and find ↵ and ! in terms of
and !0 .
Given x = Ae ↵t cos !t we di↵erentiate it using the product rule to find its first and second derivatives dx = !Ae ↵t sin(!t) ↵Ae ↵t cos(!t) dt d2 x = ! 2 Ae ↵t cos(!t) + ↵!Ae ↵t sin(!t) + ↵!Ae 2 dt = (↵2 ! 2 )Ae ↵t cos(!t) + 2↵!Ae ↵t sin(!t)
↵t
sin(!t) + ↵2 Ae
↵t
cos(!t)
To show its a solution we sub these derivatives back into the original equation to obtain (↵2 ! 2 )Ae ↵t cos(!t) + 2↵!Ae + !02 Ae ↵t cos(!t) = 0 (↵2
!2
↵ + !02 )Ae
↵t
↵t
sin(!t)
!Ae
cos(!t) + (2↵!
↵t
sin(!t)
!)Ae
↵t
↵Ae
↵t
sin(!t) = 0
cos(!t)
For this to be true for all values of t the coefficients of the cosine and sine functions must be 0. This means we must have that: ↵2
! 2 + !02 2↵!
↵=0 ↵! = 0
(1) (2)
Looking first at (2): 2↵!
↵! = 0 ↵=
2
And using this in (1): ↵2 2
4
! 2 + !02
↵=0
! 2 + !02
2
=0
2
! 2 + !02
=0
4
) ! 2 = !02 r !=
So x = Ae
↵t
cos !t is a solution when ↵ =
2
!02
2
4 2
4 q and ! = !02
2
4
.
* 4. An object of mass 0.2 kg is hung from a spring whose spring constant is 80 N/m. The object is subject to a resistive force given by bv, where v is its velocity in meters per second. (a) Set up the di↵erential equation of motion for free oscillations of the system. p (b) If the damped frequency is 3/2 of the undamped frequency, what is the value of the constant b? (c) What is the Q of the system? (a) We have a damped oscillator where the damping term is bv. So the sum of the forces is F = kx bv. By Newton’s second law the sum of the forces must equal ma so we have ma = kx bv. ma + bv + kx = 0 dx dx m 2 + b + kx = 0 dt dt d2 x b dx k + + x=0 dt2 m dt m d2 x dx + + !02 x = 0 2 dt dt 2
p where = b/m and !0 = k/m. Using our values for m and k from the question we can write: b b = m 0.2r r p k 80 !0 = = = 400 = 20 m 0.2 =
(b) !0 is the undamped angular p frequency and ! is the damped angular frequency. We have the relation ! = 3/2!0 . From the previous question we know that ! 2 = !02 3 2 !0 = !02 4
2
2
/4. Therefore
/4
b m ) b = !0 m = 20(0.2) = 4 !0 =
=
(c) Q is given by the ratio of the constants !0 / .
Q=
!0
=
20 =1 20
5. Many oscillatory systems, although the loss or dissipation mechanism is not analogous to viscous damping, show an exponential decrease in their stored average energy with time E¯ = E¯0 e t . A Q for such oscillators may be defined using the definition Q = !0 , where !0 is the natural angular frequency. (a) When the note ”middle C” on the piano is struck, its energy of oscillation decreases to one half its initial value in about 1 sec. The frequency of middle C is 256 Hz. What is the Q of the system? (b) If the note an octave higher (512 Hz) takes about the same time for its energy to decay, what is its Q? (c) A free, damped harmonic oscillator, consisting of a mass m = 0.1kg moving in a viscous liquid of damped coefficient b (Fviscous = bv), and attached to a spring of spring constant k = 0.9N m 1 , is observed as it performs oscillatory motion. Its average energy decays to 1e of its initial value in 4 sec. What is the Q of the oscillator? What is the value of b? (a) We know that E¯ = E¯0 e Using this:
t
. E¯ = E¯0 e
t
after 1 second; t = 1 we have: E¯0 = E¯0 e 2 1 =e 2
(1)
✓ ◆ 1 ln = 2 ln 2 1 = ln (2) = = ln(2) = 0.69
)
To find Q we also need to find !0 !0 =
2⇡ = 2⇡f = 2⇡(256) = 512⇡ T
Now we can use Q = !0/ Q=
!0
=
512⇡ = 2331.15 0.69
(b) Our change in energy is the same so we still have
= 0.69
!00 = 2⇡f 0 = 2⇡(512) = 2⇡(256)(2) = 2⇡(2f ) = 2(2⇡f ) where f is the frequency from part (a), this gives us: !00 = 2!0 Using this we can find our Q0 for this note: Q0 =
!00
=2·
!0
= 2Q = 2(1331.15) = 4662.3
(c) To find b, we know that E¯ = E¯0 e Using this: E¯ = E¯0 e after 1 second; t = 1 we have: E¯0
t
.
t
1
= E¯0 e
(4)
e )1=4 1 = 4 Using
= b/m: 1 b = 4 m m 0.1 )b= = = 0.025 4 4
To find Q: Q=
!0
=
r
k 1 = m
r
p 0.9 1 = 94 = 3(4) = 12 0.1 1/4
6. A U-tube has vertical arms of radii r and 2r, connected by a horizontal tube of length ` whose radius increases linearly from r to 2r. The U-tube contains liquid up to a height h in each arm. The liquid is set oscillating, and at a given instant the liquid in the narrower arm is a distance y above the equilibrium level.
*(a) Show that the potential energy of the liquid is given by U = 58 g⇢⇡r2 y 2 . *(b) Show that the kinetic energy of a small slice of liquid in the horizontal arm (see the diagram) is given by ✓ ◆2 1 ⇡r2 dx dy dK = ⇢ . 2 (1 + x/`)2 dt (Note that, if liquid is not to pile up anywhere, the product velocity ⇥ cross section must have the same value everywhere along the tube.)
(c) Using the result of part (b), show that the total kinetic energy of all the moving liquid is given by ✓ ◆2 1 5 dy 2 K = ⇢⇡r (` + h) . 4 2 dt (Ignore any nastiness at the corners.) (d) From (a) and (c) calculate the period of oscillations of ` = 5h/2. (a) Potential energy is U = mgh.
Narrow column (I) gains P.E while wider column (II) loses P.E. Since radii are di↵erent to find height h that liquid is lifted we must find h on column (I) and h on column (II), then their average. Since no liquid is displaced the volumes must equal. Let x be the height the liquid goes down in column (II). The volumes are VI = ⇡r2 y and VII = 4⇡r2 x. Equating the volumes we find that x = (1/4)y and therefore the average height is: ✓ ◆ 1 1 5 average = y+y = y 2 4 8 Subbing this into the expression for the P.E gives U = mg(5/8)y. We now find the mass of this piece of liquid. mass=density ⇥ cross sectional area, so m = ⇢⇡r2 y and the potential energy is given by 5 U = g⇢⇡r2 y 2 8
(b) Told velocity ⇥ cross section = constant everywhere. To find the cross section we need the radius at each point. r(x = 0) = r0 = r r(x > 0) = rx = r(1 + x/`)
A(x = 0) = A0 = ⇡r2 A(x > 0) = Ax = ⇡r2 (1 + x/`)
) )
Note that at x = 0, r = r and at x = `, r = 2r as desired. Product velocity by cross section is constant implies that: A0
dy dx = Ax dt dt
)
dx A0 dy = dt Ax dt
Kinetic energy is (1/2)mv 2 and the mass of dx = density ⇥ volume = ⇢Ax dx, hence the kinetic energy of dx is: ✓ ◆2 ✓ ◆2 ✓ ◆2 ✓ ◆2 1 dx 1 A0 dy 1 A20 dy dK = ⇢Ax dx = ⇢Ax dx = ⇢dx 2 dt 2 Ax dt 2 Ax dt ✓ ◆ ✓ ◆ 2 2 1 (⇡r2 )2 dy 1 ⇢dx⇡r2 dy = ⇢dx 2 = 2 2 2 ⇡r (1 + x/`) dt 2 (1 + x/`) dt (c) Use kinetic energy is (1/2)mv 2 for columns (I) and (II) and integrate answer of part (b) for x = 0 to x = ` to find kinetic energy of liquid in horizontal arm. For column (I), x = 0 so velocity all in y direction. ✓ ◆2 1 2 1 dy 2 KEI = mv = ⇢⇡r h 2 2 dt Similarly for column (II) KEII
1 1 = mv 2 = ⇢4⇡r2 h 2 2
✓
dx dt
◆2
,
but at this column x = ` and therefore dx A0 dy ⇡r2 dy 1 dy = = 2 = . dt Ax dt ⇡r (1 + x/`)2 dt 4 dt So filling this back in we get KEII
1 1 = mv 2 = ⇢4⇡r2 h 2 2
✓ ◆2 ✓ ◆2 1 dy . 4 dt
To find KE of liquid in horizontal arm we integrate answer part (b) from x = 0 to x = `, that is ✓ ◆2 Z ` ✓ ◆2 Z ` 1 dy dx 1 dy 2 2` KEIII = dK = ⇢⇡r = ⇢⇡r 2 2 dt 2 2 dt 0 0 (1 + x/`) Total KE is KEI + KEII + KEIII , which is ✓ ◆2 ✓ ◆2 ✓ ◆2 ✓ ◆2 1 dy 1 1 dy 1 dy 2 2 2` KE = ⇢⇡r h + ⇢4⇡r h + ⇢⇡r 2 dt 2 4 dt 2 2 dt ✓ ◆ ✓ ◆2 1 5h dy = ⇢⇡r2 ` + 4 2 dt
(d) Use answers to parts (a) and (c) to find the total energy, the sum of potential and kinetic. 1 E = KE + P E = m 2
✓
dx dt
◆2
1 + kx2 2
We have from parts (a) and (c) that ✓ ◆ ✓ ◆2 1 5h dy 5 2 E = KE + P E = ⇢⇡r ` + + g⇢⇡r2 y 2 4 2 dt 8 " ✓ ◆ ✓ ◆2 # 1 1 5h dy 1 5 2 = ⇢⇡r ` + + g⇢⇡r2 y 2 2 2 2 dt 2 4 So clearly m = (1/2)⇢⇡r2 (` + 5h/2) and k = (5/4)g⇢⇡r2 , so s r r k 5g/4 g != = = m (1/2)(` + 5h/2) 2h when ` = 5h/2. This also means that the period T is s 2⇡ 2h T = = 2⇡ . ! g
Vibrations and Waves MP205, Assignment 5 Solutions 1. Consider how to solve the steady-state motion of a forced oscillator if the driving force is of the form F = F0 sin(!t) instead of F0 cos(!t). For a forced oscillator with driving force F0 sin(!t), it’s equation of motion is m
d2 x + kx = F0 sin(!t) dt2
To solve we assume the solution is of the form x = C sin(!t) and solve for the constant C. This means dx/dt = !C cos(!t) and d2 x/dt2 = ! 2 C sin(!t). Filling these into the equation of motion (EOM) we find that m! 2 C sin(!t) + kC sin(!t) = F0 sin(!t). Equating coefficients gives F0 F0 F0 m C= = k = 2m 2 2 m! 2 + k !0 ! ! m q k since !0 = m . We want to express x in terms of a sinusoidal vibration having an amplitude A, by defining a positive quantity , and a phase ↵ at t = 0: x = A cos(!t + ↵) We have:
⇣
⇡⌘ x = C sin(!t) = C cos !t + 2 ⇣ ⇡⌘ = |C| cos !t + ⇣ 2 ⇡⌘ = |C|( 1) cos !t + 2
for C > 0 ) !0 > ! for C < 0 ) ! > !0
We set A = |C| to be our (positive) amplitude, and use the fact that cos(x) = cos(x + ⇡): ⇣ ⇡⌘ x = A cos !t + for !0 > ! 2 ✓ ◆ ⇣ ⌘ ⇡ 3⇡ = A cos !t + + ⇡ = A cos !t + for ! > !0 2 2 This gives us:
x = A cos (!t + ↵) Where: A = |C| C=
!02 ⇡ ↵= 2 3⇡ ↵= 2
F0 m
!2 for !0 > ! for ! > !0
2. An object of mass 0.2 kg is hung from a spring whose spring constant is 80 Nm 1 . The body is subject to a resistive force given by bv, where v is its velocity in ms 1 and b = 4 Nm 1 sec. (a) Set up the di↵erential equation of motion for free oscillations of the system, and find the period of such oscillations. (b) The object is subjected to a sinusoidal driving force given by F (t) = F0 sin(!t), where F0 = 2 N and ! = 30 sec 1 . In the steady state, what is the amplitude of the forced oscillation? (c) What is the mean power input? (d) Show that the energy dissipated against the resistive force in one cycle is 0.063J (a) We know the mass is subject to a resistive force the spring kx: F = kx bv ma + bv + kx = 0 0.2a + 4v + 80x = 0 a + 20v + 400x = 0 2 dx dx + 20 + 400x = 0 2 dt dt Comparing this to the general form: d2 x dx + + !02 x = 0 dt2 dt We can read o↵ values for and !0 : = 20 2 !0 = 400
bv, as well as a force due to
!0 = 20
To obtain the period of oscillation we require !: 2 400 ! 2 = !02 = 400 = 400 4 4 p p ! = 300 = 10 3 2⇡ 2⇡ ⇡ T = = p = p ⇡ 0.36s ! 10 3 5 3
100 = 3
(b) For a system with the equation of motion: d2 x dx F0 + + !02 x = cos !t 2 dt dt m The amplitude is given by: A(!) = p 2 (!0
F0 m ! 2 )2
+ ( !)2
From (a) we know that: !0 = 20, m = 0.2 and = 20. We’re told here that ! = 30 and F0 = 2, using this in our equation for the amplitude gives: A(!) = p ((20)2 =p ((400
2 0.2 (30)2 )2
+ ((20)(30))2
10 900)2 + (600)2 10
=p (( 500)2 + 360, 000 10 =p 250, 000 + 360, 000 10 =p 610, 000 = 0.0128m (c) The mean power is given by: F 2 !0 P¯ = 0 ⇣ 2kQ !0
1 ! !0
!
⌘2
Using the same values we used in (b), Q = (20)2 (0.2) = 80 we get: (2)2 (20) P¯ = 2(80)(1) =
4(20) 160
+
!0
1 Q2 20 20
=
= 1 and k = !02 m =
1 20 30
30 2 20
+
1 1
1 5 2 6
+1
80 1 25 160 36 + 1 1 1 = 61 2 36 1 36 = 2 61 18 = 61 ⇡ 0.3W =
(d) The energy lost per cycle is given by: 2⇡ 2⇡ E = P¯ T = P¯ = (0.3) = 0.063J ! 30 * 3. A block of mass m is connected to a spring, the other end of which is fixed. There is also a viscous damping mechanism. The following observations have been made on the system: (1) If the block is pushed horizontally with a force equal to mg, the static compression of the spring is equal to h.
(2) The viscous resistive force is equal to mg if the block moves with a certain known speed u. (a) For this complete system (including both spring and damper) write the di↵erential equation governing horizontal oscillations of the mass in terms of m,g, h and u. (a) (1) tells us: kx
x=h
= mg
kh = mg mg k= h (2) tells us: bv
v=u
= mg
bu = mg mg b= u Using this in: ma + bv + kx = 0 dx dx m 2 + b + kx = 0 dt dt 2
gives:
d2 x mg dx mg + + x=0 dt2 u dt h d2 x g dx g + + x=0 dt2 u dt h Comparing this with d2 x dx + + !02 x = 0 2 dt dt gives us values for and !0 : g = u g !02 = h m
) !0 =
p Answer the following for the case that u = 3 gh
r
g h
(b) What is the angular frequency of the damped oscillations? q (c) After what time, expressed as a multiple of hg , is the energy down by a factor 1e ?
(d) What is the Q of this oscillator? p Using u = 3 gh gives us: g g 1 = = p = u 3 3 gh g !02 = h
r
g h
2
(b) To find the angular frequency we use: ! 2 = !02
4
.
2
! 2 = !02
4 g 1 g = h 9 4h g g = h 36h 35g = 36h r 35g != 36h
(c) The energy decreases according to E(t) = E0 e
t
t
E(t) = E0 e we need to find a ⌧ st: 1
E(⌧ ) = E0 e
= E0 e
⌧
this tells us 1= ⌧ 1 ⌧= s =3
h g
So the time taken for the energy to decrease a factor of (d) Q=
1 e
is t = 3
q
h s. g
!0 pg
= 1 phg 3
=
h
1 1 3
=3 4. A mass m is subject to a resistive force force.
bv but no springlike restoring
* (a) Show that its displacement as a function of time is of the form: x = C v0 e t where = mb (b) At t 0 the mass is at rest at x = 0. At this instant a driving force F = F0 cos !t is switched on. Find the values of A and in the steady-state solution x = A cos(!t ) (c) Write down the general solution [The sum of parts (a) and (b)] and find the values of C and v0 from the conditions that x 0 and dx =0 dt at t = 0
(a) F = ma =
bv bv b v= m
a=
v
Now, we know that a = dv : dt dv = v dt 1 dv = dt v Integrating both sides gives Z Z Z 1 dv = dt = dt v ln(v) = t + D where D is a constant v=e
t+D
v(t) = eD e t Let v0 be the inital velocity at the time 0: v(0) = eD e0 = v0 ) e D = v0 This gives us a final expression for v v(t) = v0 e t To get an expression for x we use the fact that v = dx =v dt dx = v0 e t dt dx = v0 e t dt Integrating both sides gives Z Z Z t dx = v0 e dt = v0 e x=
)x=C
v0
e
v0
t
+C
e
t
t
dx dt
dt
where C is a constant
(b) ma + bv = F0 cos !t dx dx m 2 +b = F0 cos !t dt dt d2 x b dx F0 + = cos !t 2 dt m dt m 2
Looking at the steady state solution: x = A cos(!t expressions for A and .
), we want to obtain
Going to the complex-exponential method; our basic equation becomes: d2 z b dz F0 i!t + = e 2 dt m dt m We assume the solution: z = Aei(!t z = Aei(!t
)
, with x = Re(x).
delta)
dz = i!Aei(!t ) dt d2 z = ! 2 Aei(!t dt2
)
Using these in our EOM: d2 z b dz + = 2 dt m dt b ! 2 Aei(!t ) + i!Aei(!t ) = m b ! 2 Ae i + i!Ae i = m b ! 2 A + i!A = m b ! 2 A + i!A = m b ! 2 A + i!A = m
F0 i!t e m F0 i!t e m F0 m F0 i e m F0 (cos( ) + i sin( )) m F0 F0 cos( ) + i sin( ) m m
Equating the real and imaginary parts: !2A = ) cos( ) = b !A = m ) sin( ) = tan( ) = =
F0 cos( ) m m 2 ! A F0 F0 sin( ) m b !A F0 b !A sin( ) F0 = m 2 cos( ) ! A F0 b b 1 = = m! m! !
We can easily work out values for cos( ) and sin( ):
p
cos( ) =
sin( ) = p
2
! 2
+ !2
+ !2
Note: we’ve chosen the signs here to ensure we have a positive value for A. Using this to get an expression for A: F0 cos( ) m F0 A= cos( ) m! 2 F0 ! p = 2 2 + !2 m! F p 0 = 2 + !2 m!
!2A =
v0
e t + A cos(!t dx (0) = 0 dt v0 x(t) = C e t + A cos(!t
(c) The general solution is give by x = C We have the initial conditions: x(0) =
x(0) = C C
v0
v0
e0 + A cos(0
)
) )=0
+ A cos( ) = 0
dx (t) = v0 e t !A sin(!t ) dt dx (0) = v0 e0 !A sin(0 )=0 dt v0 + !A sin( ) = 0 v0 = !A sin( ) ! =
!A
=
Ap
= =
p
2
+ !2
!
2
+ !2 F !A p 0 p 2 2 2 m! +! + !2 F0 m( 2 + ! 2 )
And to find C: C
v0
+ A cos( ) = 0 C=
v0
A cos( ) m(
=
F0 2 +! 2 )
+
F p 0
+ !2 F0 F0 + 2 2 2 m( + ! ) m( + ! 2 )
=
m!
2
p
! 2
+ !2
=0 * 5. The graph shows the power resonance curve of a certain mechanical system when driven by a force F0 sin(!t), where F0 = constant and ! is variable.
(a) Find the numerical values of !0 and Q for this system. (b) The driving force is turned o↵. After how many cycles of free oscillation is the energy of the system down to 1/e5 of its initial value? (e = 2.718) (To a good approximation, the period of free oscillation can be set equal to 2⇡/!0 .) (a) Here we use the fact that with width of the power-resonance curve at halfheight ⇡ !0 = Q Q=
⇡2 !0 40 = = 20 2 2
(b) The energy decreases according to E(t) = E0 e
t
t
E(t) = E0 e we need to find a ⌧ st: E(⌧ ) = E0 e
5
= E0 e
this tells us 5= ⌧ 5 5 ⌧ = = = 2.5 2
⌧
The time taken to complete one cycle is 2⇡ n cycles is n omega : 0
2⇡ , omega0
⌧ = 2.5 = n )n
so the time taken to complete
2⇡ omega0
2⇡ = 2.5 omega0 2⇡ n = 2.5 40 n(0.16) = 2.5 2.5 n= = 15.6 ⇡ 16 0.16
6. The figure shows the mean power input P¯ as a function of driving frequency for a mass on a spring with damping. (Driving force = F0 sin(!t), where F0 is held constant and ! is varied.) The Q is high enough so that the mean power input, which is maximum at !0 , falls to half-maximum at the frequencies 0.98!0 and 1.02!0 .
(a) What is the numerical value of Q? (b) If the driving force is removed, the energy decreases according to the equation E = E0 e t . What is the value of ? (c) If the driving force is removed, what fraction of the energy is lost per cycle? (a) !0 = ⇡ width of the power-resonance curve at half-height Q !0 = 1.02!0 0.98!0 Q = 0.04!0 !0 Q= 0.04!0 1 = = 25 0.04 (b) From (a) we can just write down the value of : = 0.04!0
(c) The energy decreases according to the equation E = E0 e of energy lost is EE0 :
t
, so the fraction
E E0 e t = E0 E0 =e t = e 0.04!0 t The time taken for one cycle is the perios T = lost per cycle is ⇣ ⌘ 2⇡ E 0.04!0 ! 0 =e E0 = e 0.08⇡ s
2⇡ , !0
so the fraction of energy
A new system is made in which the spring constant is doubled, but the mass and the viscous medium are unchanged, and the same driving force F0 sin(!t) is applied. In terms of the corresponding quantities for the original system, find the values of the following: (d) (e) (f ) (g)
The The The The
new resonant frequency !00 . new quality factor Q0 . maximum mean power input P¯m0 . total energy of the system at resonance, E00 .
(d) For the original system: !0 = For the new system: !00 =
r
k m
r
k0 = m
r
2k p = 2 m
r
p k = 2!0 m
(e) For the original system: Q=
!0
For the new system: Q0 =
!00
=
p !0 p 2 = 2Q
(f) For the original system: QF02 P¯max = 2m!0 For the new system: p 0 2 Q F 2QF02 QF02 0 0 p P¯max = = = = P¯max 2m!00 2m!0 2m 2!0 (g) Originally we had E = E0 e t , therefore E0 = Ee t . As there is no k dependence here we see that E00 = Ee t = E0 .
Vibrations and Waves MP205, Assignment 6 Solutions * 1. Two identical pendulums are connected by a light coupling spring. Each pendulum has a length of 0.4 m, and they are at a place where g = 9.8 ms 2 . With the coupling spring connected, one pendulum is clamped and the period of the other is found to be 1.25 sec exactly. (a) With neither pendulum clamped, what are the periods of the two normal modes? (b) What is the time interval between successive maximum possible amplitudes of one pendulum after one pendulum is drawn aside and released? (I’ll go through this first question in quite some detail ) (a) Our system is given by:
We can write a general equation for A, by looking at the case where we move A a distance xA , and B a distance xB and seeing what forces a↵ect A as a result.
There are two forces a↵ecting A here: • restoring force due to the pendulum a= ma = F =
!02 x m!02 x m!02 x
• restoring force due to the spring F =
kx
where x is the change in spring length: xA
xB
So the total force on A is:
setting !c =
q
FA = d 2 xA m 2 = dt d 2 xA = dt2 k m
m!02 xA
k(xA
xB )
m!02 xA
k(xA
xB )
!02 xA
k (xA m
xB )
we get:
d 2 xA + !02 xA + !c2 (xA dt2 Similarly, the equation of motion for B is: d 2 xB + !02 xB + !c2 (xB dt2
xB ) = 0
xA ) = 0
We’re told that when B is clamped (xB = 0), that the period of A is TA = 1.25s. 2⇡ !A 2⇡ 2⇡ !A = = = 5.03 TA 1.25 !A2 = (5.03)2 = 25.27 TA =
Looking at this system:
d 2 xA + !02 xA + !c2 (xA xB ) = 0 dt2 d 2 xA + !02 xA + !c2 (xA 0) = 0 dt2 d 2 xA + !02 xA + !c2 xA = 0 2 dt d 2 xA + (!02 + !c2 )xA = 0 dt2 q ) !A =
!02 + !c2
For a simple pendulum of length l, the angular frequency !0 is given by: r g !0 = l g !02 = l So in our case: 9.8 !02 = = 24.5 0.4
Using this in our expression for !A gives: q !A = !02 + !c2 p !A = 24.5 + !c2 !A2 = 24.5 + !c2 25.27 = 24.5 + !c2 !c2 = 25.27 24.5 = 0.77 !c = 0.88 To find the normal modes we let xA = C cos(!t) and xB = D cos(!t) and obtain an expression for ! that satisfies both equations of motion.: xA = C cos(!t) xB = D cos(!t)
d 2 xA = ! 2 C cos(!t) 2 dt d 2 xB = ! 2 D cos(!t) dt2
Using this in our equations of motion: First for A: d 2 xA + !02 xA + !c2 (xA xB ) = 0 2 dt ! 2 C cos(!t) + !02 C cos(!t) + !c2 (C cos(!t) D cos(!t)) = 0 ! 2 C + !02 C + !c2 (C D) = 0 C( ! 2 + !02 + !c2 ) !c2 D = 0 C !c2 = D ! 2 + !02 + !c2 And for B: d 2 xB + !02 xB + !c2 (xB xA ) = 0 2 dt ! 2 D cos(!t) + !02 D cos(!t) + !c2 (D cos(!t) C cos(!t)) = 0 ! 2 D + !02 D + !c2 (D C) = 0 D( ! 2 + !02 + !c2 ) !c2 C = 0 C ! 2 + !02 + !c2 = D !c2
Combining these results: C C ) = D D !c2 ! 2 + !02 + !c2 = ! 2 + !02 + !c2 !c2 (!c2 )2 = ( ! 2 + !02 + !c2 )2 ±!c2 = ! 2 + !02 + !c2 ! 2 = !02 + !c2 ⌥ !c2 ) ! 20 = !02 + !c2 + !c2 = !02 + 2!c2 q 0 ! = !02 + 2!c2 p p p = 24.5 + 2(0.77) = 24.5 + 2(0.77) = 26.04 = 5.1 2⇡ 2⇡ T0 = 0 = = 1.23s ! 5.1 ) ! 200 = !02 + !c2 !c2 = !02 p ! 00 = !0 = 24.5 = 4.95 2⇡ 2⇡ T 00 = 00 = = 1.27s ! 4.95 Note: normal modes for this system are given by the following two cases: (i) The case where A and B are pulled apart the same distance: xA = xB .
(ii) The case where A and B are pulled in the same direction the same distance: xA = xB .
Looking at (i), and the equation of motion for A: d 2 xA + !02 xA + !c2 (xA xB ) = 0 dt2 d 2 xA + !02 xA + !c2 (xA + xA ) = 0 dt2 d 2 xA + !02 xA + 2!c2 xA = 0 dt2 d 2 xA + (!02 + 2!c2 )xA = 0 dt2 q ) !0 =
!02 + 2!c2
Looking at (ii), and the equation of motion for A: d 2 xA + !02 xA + !c2 (xA xB ) = 0 dt2 d 2 xA + !02 xA + !c2 (xA xA ) = 0 2 dt d 2 xA + !02 xA = 0 dt2 ) ! 00 = !0 (b) Starting with our general equations of motion: d 2 xA + !02 xA + !c2 (xA dt2 d 2 xB + !02 xB + !c2 (xB 2 dt
xB ) = 0 xA ) = 0
Adding these equations: d2 (xA + xB ) + !02 (xA + xB ) = 0 dt2 and setting q1 = xA + xB we get: d 2 q1 + !02 q1 = 0 dt2 so we can write q1 as: q1 = C cos(!0 t) = C cos(4.95t) Subtracting these equations: d2 (xA xB ) + !02 (xA xB ) + 2!c2 (xA xB ) = 0 dt2 d2 (xA xB ) + (!02 + 2!c2 )(xA xB ) = 0 2 dt p using ! 0 = !02 + 2!c2 and q2 = xA xB we get: d 2 q2 + ! 02 q2 = 0 2 dt so we can write q2 as: q2 = D cos(! 0 t) = D cos(5.1t) We can write xA and xB in terms of q1 and q2 . q1 = x A + x B q2 = x A x B q1 + q2 = 2xA 1 xA = (q1 + q2 ) 2
q1 = x A + x B q2 = x A + x B q1
q2 = 2xB 1 xB = (q1 q2 ) 2
This gives us: 1 xA (t) = (q1 + q2 ) 2 1 = (C cos(4.95t) + D cos(5.1t)) 2 1 xB (t) = (q1 q2 ) 2 1 = (C cos(4.95t) D cos(5.1t)) 2 These are the general solutions for xA and xB . To go any further we need initial conditions. In this case, we’re told one pendulum is drawn aside and released. This corresponds to one pendulum having displacement 0 at t = 0, and one pendulum having some displacement (say A0 ) at t = 0. Lets assume A is drawn aside, so at t = 0: xA (0) = A0 1 (C cos(0) + D cos(0)) = A0 2 C + D = 2A0 xB (0) = 0 1 (C cos(0) D cos(0)) = 0 2 C D=0 C=D Filling this into xA (0) we get: C + C = 2A0 C = A0 D = C = A0 . This gives us our equations of motion: 1 (A0 cos(4.95t) A0 cos(5.1t)) 2 A0 = (cos(4.95t) cos(5.1t)) 2 We can use the trigonometric identity cos A cos B = 2 sin A+B sin 2 ✓ ◆ ✓ ◆ (4.95 + 5.1) 5.1 4.95 = A0 sin t sin t 2 2 = A0 sin (5t) sin (0.08t) 1 xB (t) = (A0 cos(4.95t) + A0 cos(5.1t)) 2 A0 = (cos(4.95t) + cos(5.1t)) 2 xA (t) =
A B 2
We can use the trigonometric identity cos A + cos B = 2 cos A+B cos 2 ✓ ◆ ✓ ◆ (4.95 + 5.1) 5.1 4.95 = A0 cos t cos t 2 2 = A0 cos (5t) cos (0.08t) ) xA (t) = A0 sin (5t) sin (0.08t) ) xB (t) = A0 cos (5t) cos (0.08t)
A B 2
We’re asked to find the time interval between maximum possible amplitudes of one pendulum in this system. Looking at xB we can note that this is simply a beat equation. If we plot xB against time we can see its motion:
Where the distance between peaks is the time interval between successive maximum possible amplitudes. 0 If we compare this to the motion of cos !0 +! t = cos(5t) 2
and the motion of cos
!0 ! 0 t 2
= cos(0.08t)
We can see that the time interval between successive peaks is half the period of the slow oscillating motion, which is given by: T =
2⇡ |!0 ! 0 | 2
=
4⇡ |!0
!0|
So the time we require is the beat period: 1 2⇡ 2⇡ = T0 = T = = 39.3s 2 |!0 ! 0 | |2(0.08)|
2. Two harmonic oscillators A and B, of mass m and spring constants kA and kB , respectively, are coupled together by a spring of spring constant kC . Show that the normal frequencies satisfy: s✓ ◆2 kA + kB kA k B 2 m! = + kC ± + kC2 2 2
Our equations of motion are: d 2 x A kA + xA + dt2 m d 2 x B kB + xB + dt2 m
kC (xA m kC (xB m
xB ) = 0 xA ) = 0
We want to find ! such that xA = C cos(!t) and xB = D cos(!t) are solutions to these equations. kA kC C cos(!t) + (C cos(!t) D cos(!t)) = 0 m m m! 2 C + kA C + kC (C D) = 0 D m! 2 + kA + kC = C kC k k B C ! 2 D cos(!t) + D cos(!t) + (D cos(!t) C cos(!t)) = 0 m m m! 2 D + kB D + kC (D C) = 0 D kC = 2 C m! + kB + kC ! 2 C cos(!t) +
Combining these: m! 2 + kA + kC kC = 2 kC m! + kB + kC 2 ( m! + kA + kC )( m! 2 + kB + kC ) = kC2 (m! 2 )2 + kA kB + kA kC + kB kC + kC2 m! 2 kA m! 2 kB 2m! 2 kC = kC2 (m! 2 )2 m! 2 (kA + kB + 2kC ) + (kA kB + kA kC + kB kC ) = 0
This is just a quadratic equation for m! 2 , so we can solve it using the quadratic formula: p ( (k + k + 2k )) ± (kA + kB + 2kC )2 4(1)(kA kB + kA kC + kB kC ) A B C m! 2 = 2(1) p 2 2 2 kA + kB + 2kC ± kA + kB + 4kC + 2kA kB + 4kA kC + 4kB kC 4kA kB 4kA kC = 2 p 2 kA2 + kB 2kA kB + 4kC2 kA + kB = + kC ± 2 2 r kA + kB (kA kB )2 + 4kC2 = + kC ± 2 4 s✓ ◆2 kA + kB kA kB 4 = + kC ± + kC2 2 2 As required. 3. Two equal masses on an e↵ectively frictionless horizontal air track are held between rigid supports by three identical springs, as shown. The displacements from equilibrium along the line of the springs are described by coordinates xA and xB , as shown. If either of the masses is clamped, the period T for one complete vibration of the other is 3 sec.
*(a) If both masses are free, what are the periods of the two normal modes of the system?
(a) Our equations of motion are:
Setting !0 =
q
d 2 xA k k + xA + (xA 2 dt m m d 2 xA k + (2xA dt2 m d 2 xB k k + xB + (xB 2 dt m m d 2 xB k + (2xB dt2 m
xB ) = 0 xB ) = 0 xA ) = 0 xA ) = 0
k : m
d 2 xA + !02 (2xA dt2 d 2 xB + !02 (2xB dt2
xB ) = 0 xA ) = 0
4kB kC
We’re told that when one mass is clamped, the period of the other is T = 3s. Lets clamp B, so xB = 0. Using this in the equation of motion for A we obtain: d 2 xA + !02 (2xA xB ) = 0 dt2 d 2 xA + 2!02 xA = 0 dt2 p ) ! = 2!0
We know the period of this system is T = 3s. 2⇡ T = =3 ! 2⇡ != 3 p 2⇡ 2!0 = 3 p 2⇡ !0 = 3 To obtain the normal modes we set xA = C cos(!t) and xB = D cos(!t)and obtain an expression for ! that satisfies both equations of motion. d 2 xA + !02 (2xA xB ) = 0 dt2 ! 2 C cos(!t) + !02 (2C cos(!t) D cos(!t)) = 0 ! 2 C + !02 (2C D) = 0 C !02 = D 2!02 ! 2 d 2 xB + !02 (2xB xA ) = 0 dt2 ! 2 D cos(!t) + !02 (2D cos(!t) C cos(!t)) = 0 ! 2 D + !02 (2D C) = 0 C 2! 2 ! 2 = 0 2 D !0 Combining these: !02
2!02 ! 2 2!02 ! 2 !02 (!02 )2 = (2!02 ! 2 )2 ±!02 2!02 ! 2 ! 2 = 2!02 ⌥ !02 p 2⇡ ! 0 = !0 = 3 p 2⇡ 2⇡ T 0 = 0 = p = 3 2s 2⇡ ! 3 p ! p p p 2⇡ 2⇡ ! 00 = 3!0 = 3 = p 3 3 p 2⇡ 2⇡ T 00 = 00 = p = 6s ! p2⇡ =
3
Note: It is easy to check that these normal modes coincide with the two following cases: • ! 0 = !0 Both masses pulled a distance x in the same direction.
p • ! 0 = 3!0 Both masses pulled a distance x in opposite directions.
At t = 0, mass A is at its normal resting position and mass B is pulled aside a distance of 5 cm. The masses are released from rest at this instant. (b) Write an equation for the subsequent displacement of each mass as a function of time. (c) What length of time (in seconds) characterises the periodic transfer of the motion from B to A and back again? (b) Starting with our equations of motion: d 2 xA + !02 (2xA xB ) = 0 2 dt d 2 xB + !02 (2xB xA ) = 0 dt2 Adding our equations of motion gives: d 2 q1 + !02 q1 = 0 dt2 where q1 = xA + xB . Subtracting our equations of motion gives: d 2 q2 + 3!02 q2 = 0 2 dt where q2 = xA xB . p SO we can write q1 = C cos(!0 t) and q2 = D cos( 3!0 t). As in question 1., we can thus write xA and xB : ⌘ p 1⇣ xA (t) = C cos(!0 t) + D cos( 3!0 t) 2 ⌘ p 1⇣ xB (t) = C cos(!0 t) D cos( 3!0 t) 2
We’re told that t = 0, mass A is at its normal resting position and mass B is pulled aside a distance of 5 cm. ) xA (0) = 0, xB (0) = 0.05 1 (C + D) = 0 2 C= D 1 xB (0) = (C D) = 0.05 2 C D = 0.05 2C = 0.05 ) C = 0.025 ) D = 0.025 xA (0) =
Which gives us:
⌘ p 0.025 ⇣ cos(!0 t) cos( 3!0 t) 2 ! p !0 + 3!0 !0 = 0.025 sin sin 2 ! p (1 + 3)!0 (1 = 0.025 sin sin 2 ⌘ p 0.025 ⇣ xB (t) = cos(!0 t) + cos( 3!0 t) 2 ! p !0 + 3!0 !0 = 0.025 cos cos 2 ! p (1 + 3)!0 (1 = 0.025 cos cos 2 xA (t) =
where: !0 =
p
p
3!0
2 p
!
3)!0
2 p
3!0
2 p 2
!
!
3)!0
!
2⇡ 3
(c) Again, as in 1., this is just the beat period: T =
|(1
2⇡ p
3)!0 |
=
|(1
p 2⇡ 3 2 = 5.8s p p2⇡ = p ( 3 1) 3) 3 |
4. Two objects, A and B, each of mass m, are connected by springs as shown. The coupling spring has a spring constant kc , and the other two springs have spring constant k0 . If B is clamped, A vibrates at a frequency ⌫A of 1.81 sec 1 . The frequency ⌫1 of the lower normal mode is 1.14 sec 1 .
(a) Briefly explain why the equations of motion of A and B are given by: d 2 xA = dt2 d 2 xB m 2 = dt m
k0 x A
kc (xA
xB )
k0 x B
kc (xB
xA )
p (b) Putting !0 = k0 /m, show that the angular frequencies !1 and !2 of the normal modes are given by !1 = !0
!2 = [!02 + (2kc /m)]1/2 ,
,
and that the angular frequency of A when B is clamped (xB = 0 always) is given by !A = [!02 + (kc /m)]1/2 . (c) Using the numerical data above, calculate the expected frequency (⌫2 ) of the higher normal mode. (The observed value was 2.27 sec 1 ). (d) From the same data calculate the ratio kc /k0 of the two spring constants. (a) Looking at our system:
Looking at A: There are two forces a↵ecting A here: • restoring force due to the spring on its left F =
k0 x A
where xA is the change in spring length, and k0 is the spring constant. • restoring force due to the spring on its right F =
kC (xA
xB )
where xA xB is the change in spring length, and kC is the spring constant. So, the total force on A is (denoting the total force on A by FA ): FA =
k0 x A
kC (xA
xB )
maA =
k0 x A
kC (xA
xB )
d 2 xA = dt2
k0 x A
kC (xA
xB )
using F = ma and a =
d2 x dt2
m
As this system is totally symmetric in A and B we can write the equation of motion for B by interchanging A and B in A’s equation of motion: m
d 2 xB = dt2
k0 x B
kC (xB
xA )
(b) Setting !0 =
q
k0 m
(after dividing across by m):
d 2 xA kC = !02 xA (xA xB ) 2 dt m d 2 xB kC = !02 xB (xB xA ) 2 dt m To obtain the normal modes we set xA = C cos(!t) and xB = D cos(!t) and obtain an expression for ! that satisfies both equations of motion. kC ! 2 C cos(!t) = !02 C cos(!t) (C cos(!t) D cos(!t)) m kC ! 2 C = !02 C (C D) m kC C = 2 kmC D !0 + m ! 2 kC ! 2 D cos(!t) = !02 D cos(!t) (D cos(!t) C cos(!t)) m kC ! 2 D = !02 D (D C) m !02 + kmC ! 2 C = kC D m Combining these: !02 +
kC m kC m
!2
=
!2 +
kC m kC m
!02
kC 2 kC ) = (!02 + ! 2 )2 m m kC kC ± = !02 + !2 m m kC kC ! 2 = !02 + ⌥ m m ) !1 = !0 r 2kC ) !2 = !02 + m Note: It is easy to check that these normal modes coincide with the two following cases: • ! 0 = !0 Both masses pulled a distance x in the same direction. (
q • ! = !02 + 2kmC Both masses pulled a distance x in opposite directions. 0
When B is clamped, xB = 0, using this in the equation of motion for A: d 2 xA kC = !02 xA (xA 2 dt m d 2 xA kC = !02 xA xA 2 dt m kC d 2 xA = (!02 + )xA 2 dt m r kC ) !A = !02 + m 1
(c) We’re told that ⌫A = 1.81s
xB )
and ⌫1 = 1.14s 1 : 1 ! = T 2⇡ = 2⇡⌫ = 2⇡⌫1 = 2⇡(1.14) = 7.16 = !0 = 7.16 = 2⇡⌫A = 2⇡(1.81) = 11.37 r kC = !02 + = 11.37 m
⌫= )! !1 !1 !A !A
kC = 129.34 m kC (7.16)2 + = 129.34 m k = 129.34 m !02 +
51.27 = 78.07
Now obtaining a numerical value for !2 is simple: r 2kC !2 = !02 + m p p 2 = (7.16) + 2(78.07) = 207.41 = 14.4 ! Using this in ⌫ = 2⇡ gives us: !2 14.4 ⌫2 = = = 2.29s 1 2⇡ 2⇡ (d) We want the ratio
kC : k0
kC = k0 =
kC m k0 m kC m !02
=
78.07 = 1.52 51.27
Vibrations and Waves MP205, Assignment 7 Solutions * 1. A string of length 1 m has a fundamental node frequency of ⌫ = 5Hz. If this string is plucked transversely and is then touched at a point in the centre, what frequencies persist? For a stretched string, the permitted stationary vibrations are given by: n ⌫n = 2L
✓ ◆ 12 T = n⌫1 µ
1 ⌫1 = 2L
✓ ◆ 12 T µ
where:
where: • L is the length of the string
• T is the tension in the string
• µ is the mass per unit length µ =
m L
⌫1 is given by: ✓ ◆1 1 T 2 ⌫1 = 2L µ = 5Hz (as given in the question) So our frequencies are given by: ⌫n = 5Hz, 10Hz, 15Hz, 20Hz, 25Hz, 30Hz, ... After the string is plucked and touched at a point in the centre, we e↵ectively half the length of the string: L0 = L2 . Our new permitted frequencies are: ⌫n0
✓ ◆ 12 T µ ✓ ◆ 12 n T = L µ 2 2 ✓ ◆ 12 n T = L µ = 2⌫n = 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ... n = 2L0
The only frequencies that persist (⌫p ) are the frequencies both systems have in common: ⌫p = 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ... 2. A uniform string of length 2.5 m and mass 0.01 kg is placed under a tension 10 N.
(a) What is the frequency of its fundamental mode? (b) If the string is plucked transversely and is then touched at a point 0.5 m from one end, what frequencies persist? (a) Again, for a stretched string, the permitted stationary vibrations are given by: n ⌫n = 2L
✓ ◆ 12 T = n⌫1 µ
1 ⌫1 = 2L
✓ ◆ 12 T µ
where:
where: • L is the length of the string • T is the tension in the string • µ is the mass per unit length µ =
m L
The frequency of its fundamental mode, ⌫1 , is given by: ✓ ◆ 12 T µ ! 12 1 T = m 2L L ✓ ◆1 1 TL 2 = 2L m ✓ ◆1 1 TL 2 = 2 mL2 ✓ ◆1 1 T 2 = 2 mL ✓ ◆ 12 1 10 = 2 (0.01)(2.5) 1 1 = (400) 2 2 1 = (20) 2 = 10Hz
1 ⌫1 = 2L
(b) The frequency of the fundamental mode for a string of length L = 2.5m is 10Hz So our frequencies are given by: ⌫n = 10Hz, 20Hz, 30Hz, 40Hz, 50Hz, 60Hz, ... After the string is plucked and touched at a point 0.5 m from one end, our new length is L0 = 2m = 4L . 5
Our new permitted frequencies are: ✓ ◆ 12 T µ ✓ ◆ 12 n T = 4L µ 2 5 ✓ ◆ 12 5n T = 8L µ 5 = ⌫n 4 = 12.5Hz, 25Hz, 37.5Hz, 50Hz, 62.5Hz, 75Hz, 87.5Hz, 100Hz, ...
n ⌫n0 = 2L0
The only frequencies that persist (⌫p ) are the frequencies both systems have in common: ⌫p = 50Hz, 100Hz, 150Hz, 200Hz, 250Hz, 300Hz, ... * 3. A string of length L and total mass M is stretched to a tension T . What are the frequencies of the three lowest normal modes of oscillation of the string for transverse oscillations? Compare these frequencies with the three normal mode frequencies of three masses each of mass M/3 spaced at equal intervals on a massless string of tension T and total length L.
(a) As in the previous question we use that the normal mode frequencies are given by:
n ⌫n = 2L
s
T µ
with
m µ= L
so
1 ⌫1 = 2
r
T mL
Three lowest are ⌫A = n⌫1 with n = 1, 2, 3, that is: s s s 1 T 1 T 3 T ⌫1 = ⌫2 = ⌫3 = 2L µ L µ 2L µ (b) For N particles on a string each of mass m and a distance ` apart: ✓
Therefore
◆ n⇡ T !n = 2!0 sin with !0 = 2(N + 1) m` r ✓ ◆ !n n⇡ 1 T ⌫n = = 2⌫0 sin with ⌫0 = 2⇡ 2N + 2 2⇡ m`
We have that m = M/3 and ` = L/4 so: r r p p 1 12T 12 1 T 12 ⌫0 = = · = ⌫1 = 1.1026⌫1 2⇡ M L ⇡ 2 ML ⇡ This means that all of the normal mode frequencies are given by: ✓ ◆ n⇡ 0 ⌫n = 2(1.1026⌫1 ) sin 2N + 2 The three lowest normal mode frequencies are when n = 1, 2, 3: ⇣⇡ ⌘ 0 ⌫1 = 2(1.1026⌫1 ) sin = 0.84⌫1 ✓ 8◆ 2⇡ ⌫20 = 2(1.1026⌫1 ) sin = 1.55⌫1 8 ✓ ◆ 3⇡ 0 ⌫3 = 2(1.1026⌫1 ) sin = 2.03⌫1 8 4. A uniform rod is clamped at the center, leaving both ends free. (a) What are the natural frequencies of the rod in longitudinal vibration? (b) What is the wavelength of the nth mode? (a) We solve the wave equation for a rod fixed at the middle, that is fixed at x = 0 and free at both ends x = ±L/2. The wave equation is given by: @ 2✏ 1 @ 2✏ = @x2 v 2 @t2 with solutions of the form ✏(x, t) = f (x) cos(!t) where f (x) = A sin(!x/v). We use boundary conditions to find a solution. Fixed at x = 0 means zero displacement at x = 0 and so f (x = 0) = 0. Filling in gives A sin(0) = 0 ) 0 = 0, so satisfied. Free at x = ±L/2 means zero stress here and so ↵Y @✏/@x = 0. @✏ =0 @x ⇣ !x ⌘ df ! = cos dx v v for
must have need
⇣ !x ⌘ ! cos v v
=0 ±L/2
)
df =0 dx ✓ ◆ !L cos ± =0 2v
cos is an even function so cos( A) = cos(A) and we have: ✓ ◆ ✓ ◆ !L !L 1 2(n 1/2)⇡v cos =0 ) = n ⇡ ) !n = 2v 2v 2 L s s (2n 1)⇡ Y (2n 1) Y !n = ) ⌫n = L ⇢ 2L ⇢ since the speed for prigid bodies is given in terms of the Young’s modulus Y and the density ⇢ as v = Y /⇢.
(b) Recall that c = f
so in this notation we have v = ⌫n
n
=
n.
v 2L L = = ⌫n 2(n 1/2) (n 1/2)
5. Derive the wave equation for vibrations of an air column. Your final result should be @ 2⇠ ⇢ @ 2⇠ = @x2 K @t2 where ⇠ is the displacement from the equilibrium position, ⇢ is the mass density, and K is the elastic modulus. This derivation is almost identical to the derivation of the wave equation for longitudinal vibrations of a rod, but using K the elastic modulus instead of Y Young’s modulus.
We consider the equation of motion of a thin slice of air, which in the undisturbed state is contained between x and x + x. Then, this slice is shifted and stretched, it is pulled in opposite directions by the forces F1 and F2 . The length of the slice (originally x) has increased by ⇠. So our average stress in this case is given by K x⇠ @⇠ The stress at x is therefore K @x @⇠ @2⇠ Similarly, the stress at x + x is K @x + @x x. 2 Taking ↵ to be the cross sectional area: @⇠ @x @⇠ @ 2⇠ + K↵ 2 x F2 = ↵K @x @x @ 2⇠ F1 = K↵ 2 x @x F1 = ↵K
F2
We now apply Newton’s law to the material lying between x and x + Taking the density to be ⇢, then the mass of this area is ⇢↵ x.
x
The acceleration is given by This gives us:
@2⇠ @x2
F @ ⇠ ⇢↵ x 2 @x @ 2⇠ ⇢ 2 @x @ 2⇠ @x2 2
= ma = F2 F1 @ 2⇠ = K↵ 2 x @x 2 @ ⇠ =K 2 @x K @ 2⇠ = ⇢ @x2
as required. * 6. A laser can be made by placing a plasma tube in an optical resonant cavity formed by two highly reflecting flat mirrors, which act like rigid walls, see figure. The purpose of the plasma tube is to produce light by exciting normal modes of the cavity.
(a) What are the normal mode frequencies of the resonant cavity? (Express your answer in terms of the distance L between the mirrors and the speed of light c.) (b) Suppose that the plasma tube emits light centered at frequency ⌫0 = 5 ⇥ 1014 Hz with a spectral width ⌫, as shown in the sketch. The value of ⌫ is such that all normal modes of the cavity whose frequency is within ±1.0 ⇥ 109 Hz of ⌫0 will be excited by the plasma tube. i. How many modes will be excited if L = 1.5 m? ii. What is the largest value of L such that only one normal mode will be excited (so that the laser will have only one output frequency)? (a) We solve the wave equation for light (massless like string) fixed at both ends, that is fixed at x = 0 and at x = L. The normal mode frequencies are given by: ⌫n =
nv nc = 2L 2L
since the speed of light v = c.
(b)(i) We find the position of the modes in terms of the mode frequency. With L = 1.5 m ⌫n = nc/3 and so n = 3⌫n /c.
when
⌫n = ⌫0 :
when
⌫n = ⌫0 +
when
⌫n = ⌫0
3(5 ⇥ 1014 ) = 5, 000, 000 3 ⇥ 108 3(5 ⇥ 1014 + 1 ⇥ 109 ) ⌫: n= = 5, 000, 010 3 ⇥ 108 3(5 ⇥ 1014 1 ⇥ 109 ) ⌫: n= = 4, 999, 990 3 ⇥ 108
n=
Between ⌫0 + ⌫ and ⌫0 we have n = 5, 000, 010 5, 000, 000 = 10 modes. Between ⌫0 and ⌫0 ⌫ we have n = 5, 000, 000 4, 999, 990 = 10 modes. So when L = 1.5 m, n = 20 modes plus the mode located at ⌫n = ⌫0 gives a total of 21 modes. (b)(ii) ⌫n = nc/2L so n = 2⌫n L/c.
when
⌫n = ⌫0 :
when
⌫n = ⌫0 +
2(5 ⇥ 1014 )L = 3, 333, 333L 3 ⇥ 108 2(5 ⇥ 1014 + 1 ⇥ 109 )L ⌫: n= = 3, 333, 340L 3 ⇥ 108
n=
We want the value of L for there to be only one excited mode, at ⌫n = ⌫0 . So want di↵erence between n at ⌫0 and n at ⌫0 + ⌫=1. So we want: (3, 333, 340 3, 333, 333)L = 6.666L = 1 ) L = 0.15 m.
Vibrations and Waves MP205, Assignment 8 Solutions 1. Find the Fourier series for the following functions (0 x L). (a) y(x) = Ax(L
x)
* (b) y(x) = A sin(⇡x/L) (c) ⇢
y(x) = (a) y(x) = Ax(L
A sin(2⇡x/L), 0 x L/2 0, L/2 x L
x) y(x) =
1 X
Bn sin
⇣ n⇡x ⌘
L ⇣ n⇡x ⌘ 2 L Bn = y(x) sin dx L 0 L Z ⇣ n⇡x ⌘ 2 L = (Ax(L x)) sin dx L 0 L Z ⇣ n⇡x ⌘i 2A L h = Lx x2 sin dx L 0 L n=0
Z
Using integration by parts:
Z
Z
udv = uv
vdu
We set: u = Lx
x2
du = (L
dv = sin
2x) dx
v= v=
⇣ n⇡x ⌘ L cos
dx
n⇡x L n⇡ L L cos n⇡x L
n⇡
Using this in our integral: Z
L 0
h
Lx
x2 sin
⇣ n⇡x ⌘i L
"
dx = (Lx Z
L 0
= (L2
x2 )
L cos n⇡x L n⇡ !
n⇡x L
L cos n⇡ L2 )
Z L L + (L n⇡ 0 Z L L = (L n⇡ 0
(L
!#L 0
2x) dx
! L cos n⇡L L 0 n⇡ ⇣ ⇣ n⇡x ⌘⌘ 2x) cos dx L ⇣ ⇣ n⇡x ⌘⌘ 2x) cos dx L
Again integrating by parts: u=L
2x
du =
dv = cos
2dx
⇣ n⇡x ⌘
v= v=
L sin
dx
n⇡x L n⇡ L n⇡x L
L sin n⇡
Using this in our integral: Z
L
(L
⇣
2x) cos
0
⇣ n⇡x ⌘⌘ L
"
dx = (L Z
L 0
2x)
L sin n⇡x L n⇡ !
n⇡x L
L sin n⇡
!#L 0
( 2) dx
! ✓ ◆ L sin n⇡L L sin (0) L = (L 2L) (L 0) n⇡ n⇡ Z L ⇣ n⇡x ⌘ 2L + sin dx n⇡ 0 L " #L ✓ ◆ cos n⇡x L sin (n⇡) 2L L = (L 2L) 0+ n⇡ n⇡ n⇡ L 0 ⇣ ⌘i 2 h L 2L n⇡x =0 cos (n⇡)2 L 0 ✓ ◆ 2 2L n⇡L cos cos (0) 2 (n⇡) L 2L2 = [cos (n⇡) 1] (n⇡)2 2L2 = [( 1)n 1] 2 (n⇡) Going back to our Bn : Bn = = = =
Z ⇣ n⇡x ⌘i 2A L h 2 Lx x sin dx L 0 L ✓ ◆Z L ⇣ ⇣ n⇡x ⌘⌘ 2A L (L 2x) cos dx L n⇡ L 0 Z ⇣ ⇣ n⇡x ⌘⌘ 2A L (L 2x) cos dx n⇡ 0 L 4AL2 [( 1)n 1] 3 (n⇡)
for n even: =
4AL2 [1 (n⇡)3
=
4AL2 [ 1 (n⇡)3
1] = 0
for n odd: 1] =
8AL2 (n⇡)3
So our Fourier series is: y(x) =
1 X
Bn sin
n=0
where: Bn =
(
⇣ n⇡x ⌘ L
8AL2 (n⇡)3
n is odd n is even
0
(b) y(x) = A sin(⇡x/L) y(x) =
1 X
Bn sin
⇣ n⇡x ⌘
L ⇣ n⇡x ⌘ 2 L Bn = y(x) sin dx L 0 L Z ⇣ ⇡x ⌘⌘ ⇣ n⇡x ⌘ 2 L⇣ = A sin sin dx L 0 L L Z ⇣ n⇡x ⌘i 2A L h ⇣ ⇡x ⌘ = sin sin dx L 0 L L n=0
Z
Using the relation: 2 sin A sin B = cos(A
B)
cos(A + B)
This gives: Z ⇣ n⇡x ⌘i 2A L h ⇣ ⇡x ⌘ Bn = sin sin dx L 0 L L ◆ ✓ ◆ Z ✓ A L ⇡(1 n)x ⇡(1 + n)x = cos cos dx L 0 L L ⇣ ⌘ 3L 2 ⇣ ⇡(1 n)x ⌘ ⇡(1+n)x sin sin L L A 5 = 4 ⇡(1 n) ⇡(1+n) L L L 0 ⇣ ⌘ 3L 2 ⇣ ⇡(1 n)x ⌘ sin sin ⇡(1+n)x L L 5 = A4 ⇡(1 n) ⇡(1 + n) ⇣ ⌘ 0 2 ⇣ ⇡(1 n)L ⌘ 3 ⇡(1+n)L sin sin L L sin (0) sin (0) 5 = A4 + ⇡(1 n) ⇡(1 + n) ⇡(1 n) ⇡(1 + n) sin (⇡(1 n)) sin (⇡(1 + n)) =A 0+0 ⇡(1 n) ⇡(1 + n) sin (⇡(1 n)) sin (⇡(1 + n)) =A ⇡(1 n) ⇡(1 + n) We can see that n = 1 will require us to divide by 0, so we’ll deal with the case n = 1 seperately. Our current Bn is therefore for all n 6= 1. A sin (⇡(1 n)) sin (⇡(1 + n)) = ⇡ 1 n 1+n
Now, sin(n⇡) = 0 when n 2 Z. n 2 Z ) (1 n) 2 Z ) sin (⇡(1 n)) = 0 n 2 Z ) (1 + n) 2 Z ) sin (⇡(1 + n)) = 0 So, for n 6= 1:
A sin (⇡(1 n)) Bn = ⇡ 1 n =0
sin (⇡(1 + n)) 1+n
For n = 1, ◆ ✓ ◆ Z ✓ A L ⇡(1 1)x ⇡(1 + 1)x B1 = cos cos dx L 0 L L ✓ ◆ Z A L 2⇡x = cos (0) cos dx L 0 L ✓ ◆ Z A L 2⇡x = 1 cos dx L 0 L " #L sin 2⇡x A L = x 2⇡ L L 0 " #L 2⇡x L sin L A = x L 2⇡ 0 " # 2⇡L L sin L A L sin (0) = L 0 + L 2⇡ 2⇡ A L sin (2⇡) = L +0 L 2⇡ A = [L] L =A So our Fourier series is: y(x) =
1 X
Bn sin
n=0
where: Bn =
⇢
0 A
⇣ n⇡x ⌘ L
n 6= 1 n=1
(c) y(x) =
⇢
A sin(2⇡x/L), 0 x L/2 0, L/2 x L
y(x) =
1 X
Bn sin
⇣ n⇡x ⌘
L ⇣ n⇡x ⌘ 2 L Bn = y(x) sin dx L 0 L ✓ ◆◆ Z L✓ Z ⇣ n⇡x ⌘ ⇣ n⇡x ⌘ 2 2 2⇡x 2 L = A sin sin dx + (0) sin dx L 0 L L L L2 L ✓ ◆ Z L ⇣ n⇡x ⌘ 2A 2 2⇡x = sin sin dx L 0 L L n=0
Z
Using the relation: 2 sin A sin B = cos(A
B)
cos(A + B)
This gives: ◆ Z L ✓ ⇣ n⇡x ⌘ 2A 2 2⇡x Bn = sin sin dx L 0 L L ◆ ✓ ◆ Z L ✓ A 2 ⇡(2 n)x ⇡(2 + n)x = cos cos dx L 0 L L ⇣ ⌘ 3 L2 2 ⇣ ⇡(2 n)x ⌘ ⇡(2+n)x sin L L A sin 5 = 4 ⇡(2 n) ⇡(2+n) L L
2
= A4
sin
⇣
⇡(2 n)x L
L
⌘
sin
⇣
⇡(2+n)x L
0
⌘ 3 L2
5 ⇡(2 n) ⇡(2 + n) ⇣ ⌘0 2 ⇣ ⇡(2 n) L ⌘ 3 L ⇡(2+n) 2 2 sin sin L L sin (0) sin (0) 5 = A4 + ⇡(2 n) ⇡(2 + n) ⇡(2 n) ⇡(2 + n) ⇣ ⌘ 2 ⇣ ⇡(2 n) ⌘ 3 ⇡(2+n) sin sin 2 2 0 + 05 = A4 ⇡(2 n) ⇡(2 + n) ⇣ ⌘3 2 ⇣ ⇡(2 n) ⌘ ⇡(2+n) sin sin 2 2 5 = A4 ⇡(2 n) ⇡(2 + n)
We can see that n = 2 will require us to divide by 0, so we’ll deal with the case n = 2 seperately. Our current Bn is therefore for all n 6= 2. ⇣ ⌘3 2 ⇣ ⇡(2 n) ⌘ ⇡(2+n) sin sin 2 2 A 5 = 4 ⇡ 2 n 2+n
Now, sin(n⇡) = 0 when n 2 Z. (2 n) n 2 Z, and n even: ) 2Z 2 ◆ ✓ ⇡(2 n) ) sin =0 2 (2 + n) n 2 Z, and n even: ) 2Z 2 ✓ ◆ ⇡(2 + n) ) sin =0 2 So, for n even: Bn = 0 For n odd: sin We use sin(x) = cos x
✓
⇡(2 ± n) 2 ⇡ 2
◆
⇣
n⇡ ⌘ = sin ⇡ ± 2 ⇣
n⇡ ⇡ ⌘ = cos ⇡ ± 2 ⌘ 2 ⇣ ⇡ n⇡ = cos ± 2 ◆ ✓2 (1 ± n) = cos ⇡ 2
n is odd, so n ± 1 is even ) (1±n) is an integer. 2 ✓ ◆ (1±n) ⇡(2 ± n) sin = ( 1) 2 for n odd 2 (This is a useful identity to know) This gives us: ⇣ ⌘3 2 ⇣ ⇡(2 n) ⌘ ⇡(2+n) sin sin 2 2 A 5 Bn = 4 ⇡ 2 n 2+n " # (1 n) (1+n) A ( 1) 2 ( 1) 2 = ⇡ 2 n 2+n ⌘ ⇣ ⌘3 2 ⇣ (1 n) (1+n) (1 n) (1+n) 2 2 2 2 2 ( 1) ( 1) + n ( 1) + ( 1) A 5 = 4 ⇡ 4 n2 using ( 1)n = ( 1) ( 1)
(1+n) 2
= ( 1)
n
:
( 1 n) 2 (1 n)
= ( 1) 2 1 using ( 1)n 1 = ( 1)n : =
( 1)
(1 n) 2
Which gives us: ⌘ ⇣ 2 ⇣ (1 n) (1 n) (1 n) 2 2 + ( 1) + n ( 1) 2 A 4 2 ( 1) Bn = ⇡ 4 n2
( 1)
(1 n) 2
(1 n)
4A( 1) 2 = ⇡ (4 n2 )
⌘3 5
For n = 2, A B2 = L A = L A = L
Z Z Z
L 2
0 L 2
0 L 2
0
"
A x L " A = x L 2 =
cos
✓
⇡(2
2)x L
cos
✓
◆ 4⇡x cos (0) cos dx L ✓ ◆ 4⇡x 1 cos dx L # L2 sin 4⇡x L 4⇡ L
L sin 0
A L = L 2 A L = L 2 A = 2
⇡(2 + 2)x L
◆
dx
0
L sin 4⇡x L 4⇡
A 6L = 6 L 42
✓
◆
# L2
✓0
4⇡ ( L 2) L
4⇡
L sin (2⇡) +0 4⇡
◆
+
3
L sin (0) 7 7 4⇡ 5
So our Fourier series is: y(x) =
1 X
Bn sin
n=0
where: Bn =
8 > < > :
⇣ n⇡x ⌘
A 2
0
(1 n) 4A( 1) 2 ⇡(4 n2 )
L
n=2 n even,
n 6= 2
n odd
2. Satisfy yourself that the following equations can all be used to describe the same progressive wave: (a) y = A sin
2⇡
(x
vt)
(b) y = A sin (2⇡(kx (c) y = A sin (2⇡[(x/ )
⌫t)) (t/T )])
(d) y =
A sin (!(t
x/v))
(e) y = AIm{exp[i2⇡(kx
⌫t)]}
(a) y = A sin 2⇡ (x vt) We’ll use this equation as our base equation, and attempt to rewrite the following equations as this equation. (b) y = A sin 2⇡(kx
⌫t) y = A sin (2⇡(kx
⌫t))
Using the identites: k= ⌫=
1 v
we can rewrite our equation as: y = A sin (2⇡(kx ⌫t)) ✓ ✓✓ ◆ ⇣ v ⌘ ◆◆ 1 = A sin 2⇡ x t ◆ ✓ 2⇡ = A sin (x vt) = (a)
(c) y = A sin 2⇡[ x
(t/T )] y = A sin 2⇡[
x
(t/T )]
Using the identity: 1 ⌫ we can rewrite our equation as: y = A sin 2⇡[(x/ ) (t/T )] ✓ ◆ x t = A sin 2⇡[ 1 ] T =
x
⌫
= A sin 2⇡[ t⌫] ✓ ◆ 2⇡ = A sin (x vt) = (a) (d) y =
A sin !(t
x/v) y=
A sin !(t
x/v)
Using the identity: sin(x) = sin( x) 2⇡ v != = 2⇡⌫ = 2⇡ T
we can rewrite our equation as:
v A sin 2⇡ (t x/v) v = A sin 2⇡ (x/v t) ✓ ◆ 2⇡ = A sin (x vt)
y=
= (a) (e) y = AIm{exp[i2⇡(kx
⌫t)]} y = AIm{exp[i2⇡(kx
⌫t)]}
Using the identity: exp[i2⇡(kx ⌫t)] = cos 2⇡(kx ⌫t) + i sin 2⇡(kx Im{exp[i2⇡(kx ⌫t)]} = sin 2⇡(kx ⌫t) we can rewrite our equation as: y = A sin 2⇡(kx ⌫t) = (b) = (a)
⌫t)
3. The equation of a transverse wave traveling along a string is given by y = 0.3 sin ⇡(0.125x 25t), where y and x are in centimeters and t is in seconds. (a) Find the amplitude, wavelength, wave number, frequency, period, and velocity of the wave. (b) Find the maximum transverse speed of any particle in the string. (a) y = 0.3 sin (⇡(0.125x 25t)) = 0.3 sin (2⇡(0.0625x 12.5t)) = 0.3 sin (2⇡(0.0625x 12.5t)) So we have something of the form: y = A sin (2⇡(kx ⌫t)) or: y = 0.3 sin (2⇡(0.0625x 12.5t)) = 0.3 sin (2⇡(0.0625)(x 200t)) ✓ ◆ 2⇡ = 0.3 sin (x 200t) 16 something of this form: ✓ ◆ 2⇡ y = A sin (x vt)
Now we can read o↵ the solutions: A = 0.3cm = 16cm k = 0.0625 ⌫ = 12.5Hz 1 1 T = = = 0.08s ⌫ 12.5 v = 200cms 1 (b) y = 0.3 sin (⇡(0.125x 25t)) dy = 0.3( 25)⇡ cos (⇡(0.125x 25t)) dt = 7.5⇡ cos (⇡(0.125x 25t)) This takes its maximum value when cos (⇡(0.125x 25t)) = ±1: vmax = ⌥7.5⇡ max speed = |vmax | = 7.5⇡ms 1 = 23.56ms 1 * 4. What is the equation for a transverse wave travelling in the negative x direction with amplitude 0.003m, frequency 5 sec 1 , and speed 3000m/sec? The equation for a transverse wave travelling in the negative x direction is: ✓ ◆ 2⇡ y = A sin (x + vt) So we need values for A, , and v. From the information given we have A = 0.003m and v = 3000ms 1 , so we just need to get a value for . ⌫= )
v
v ⌫ 3000 = 5 = 600m =
Filling in these values gives: y = 0.003 sin
✓
2⇡ (x + 3000t) 600
◆
5. What is the equation for a longitudinal wave travelling in the positive x direction with amplitude 0.02m, period 1.25 sec, and speed 560m/sec? The equation for a longitudinal wave travelling in the positive x direction is: ✓ ◆ 2⇡ y = A sin (x vt)
So we need values for A, , and v. From the information given we have A = 0.02m and v = 560ms 1 , so we just need to get a value for . T = ⌫= )T = )
1 ⌫ v
v
= Tv = (1.25)(560) = 700m
Filling in these values gives: y = 0.02 sin * 6. A wave of frequency 20 sec
1
✓
2⇡ (x 700
560t)
◆
has a velocity of 80 m/sec.
(a) How far apart are two points whose displacements are 30 apart in phase? (b) At a given point, what is the phase di↵erence between two displacements occurring at times separated by 0.01 sec? (a) Two points one wavelength apart are 2⇡ apart in phase. 30 = ⇡6 rad ⌫= )
v
v ⌫ 80 = 20 = 4m =
So two points whose displacements are 2⇡ apart in phase are 4m apart. 2⇡ , 4 1 1 (2⇡) , (4) 6 6 ⇡ 4 , = 0.67m 3 6 So two points whose displacements are
⇡ 3
apart in phase are 0.67m apart.
(b) The phase di↵erence between two displacements occurring at times separated by T s are 2⇡ apart in phase. (where T is the period.) 1 ⌫ 1 = 20 = 0.05m
T =
The phase di↵erence between two displacements occurring at times separated by 0.05s are 2⇡ apart in phase. 0.05 , 2⇡ 1 1 (0.05) , (2⇡) 5 5 2⇡ 0.01 , 5 The phase di↵erence between two displacements occurring at times separated by 0.01s are 2⇡ apart in phase. 5 7. A long uniform string of mass density 0.1 kg/m is stretched with a force of 40N. One end of the string (x = 0) is oscillated transversely with an amplitude of 0.02 m and a period of 0.1 sec, so that traveling waves in the +x direction are set up. (a) What is the velocity of the waves? (b) What is their wavelength? (c) If at the driving end (x = 0) the displacement (y) at t = 0 is 0.01 m with dy/dt positive, what is the equation of the traveling waves? q q p T 40 (a) v = µ = 0.1 = 400 = 20ms 1 (b)
=
v ⌫
= vT = (20)(0.1) = 2m
(c) Starting with the usual expression for y: ✓ ◆ 2⇡ y(x, t) = 0.02 sin (x 20t) 2 y(0, 0) = 0.02 sin(0) = 0 6= 0.01 So we can see we need to include a phase to satisfy our initial conditions. ✓ ◆ 2⇡ y(x, t) = 0.02 sin (x 20t) + 2 y(0, 0) = 0.02 sin ( ) = 0.01 1 sin( ) = 2 ⇡ 5⇡ = or 6 6 1 (both satisfy sin( ) = 2 ) To work out which one we should use, we use the face that dy/dt is positive at this point: ✓ ◆ dy 2⇡ = 0.02(2⇡)( 20) cos (x 20t) + dt 2 ✓ ◆ 2⇡ = 0.8⇡ cos (x 20t) + 2 dy = 0.8⇡ cos ( ) dt x=0,t=0
So we need the value of cos ( ) that will give us a negative value: cos(
5⇡ )<0 6
) y(x, t) = 0.02 sin
✓
2⇡ (x 2
5⇡ 20t) + 6
◆
Vibrations and Waves MP205, Assignment 9 Solutions 1. Two points on a string are observed as a traveling wave passes them. The points are at x1 = 0 and x2 = 1 m. The transverse motions of the two points are found to be as follows: y1 = 0.2 sin(3⇡t) y2 = 0.2 sin(3⇡t + ⇡/8) (a) What is the frequency in Hertz? (b) What is the wavelength? (c) With what speed does the wave travel? (a) y = A sin = A sin = A sin
✓ ✓ ✓
2⇡
(x + vt)
2⇡x 2⇡x
+
◆
2⇡vt
+ 2⇡⌫t
◆
◆
Comparing this with expression above: 2⇡⌫t = 3⇡t 2⌫ = 3 ⌫ = 1.5Hz (b) 2⇡x2
=
⇡ 8
but x2 = 1 2⇡
⇡ 8 1 1 = 16 = 16m =
(c) v=⌫ = (1.5)(16) = 24ms 1 2. A symmetrical triangular pulse of maximum height 0.4m and total length 1.0m is moving in the positive x direction in a string on which the wave speed is 24ms 1 . At t = 0 the pulse is entirely located between x = 0 and x = 1m.
(a) Find an expression for the transverse velocity versus time at x = x2 = +1m. (b) Sketch the appearance of the string at t = 0sec and at t = 0.0625sec. (a) y(x, t) = = = vy = = vy
x=1
=
b3 b2 + (x vt)2 (0.4)3 (0.4)2 + (x 24t)2 0.064 0.16 + (x 24t)2 @y @t 3.072(x 24t) [0.16 + (x 24t)2 ]2 3.072(1 24t) [0.16 + (1 24t)2 ]2
(b) When t = 0 we’re told that the pulse is entirely located between x = 0 and x = 1m:
For t = 0.0625, we know the wave speed is travelling at a speed of 24ms 1 , so in a time t it moves a distance 24t. 24 ⇥ 0.025 = 1.5m
* 3. The end (x = 0) of a stretched string is moved transversely with a constant speed of 0.5ms 1 for 0.1sec (beginning at t = 0) and is returned to its normal position during the next 0.1sec, again at a constant speed. The resulting wave pulse moves at a speed of 4ms 1 .
(a) Sketch the appearance of the string at t = 0.4sec and at t = 0.5sec. (b) Draw a graph of transverse velocity against x at t = 0.4sec. (a) The max height is given by (0.5)(0.1) = 0.05m. This tells us the appearance of the wave after 0.2sec is:
The speed of the wave is 4ms 1 , so after 0.4sec the wave will have travelled 4 ⇥ 0.2 = 0.8m.
After 0.5sec the wave will have travelled 4 ⇥ 0.3 = 1.2m.
(b) b3 b2 + (x vt)2 (0.05)3 = (0.05)2 + (x 4t)2 @y 5 ⇥ 10 4 (x 4t) vy = = @t [(0.05)2 + (x 4t)2 ]2 5 ⇥ 10 4 x 8 ⇥ 10 4 vy = [x2 3.2x + 2.562]2 t=0.4 y(x, t) =
To plot this we’ll look at x = 0, vy = 0 and what happens to vy as x ! 1. ) vy =
x=0
) 5 ⇥ 10 4 x
vy = 0 x!1 x!
8 ⇥ 10 4 = (2.562)2
1
8 ⇥ 10
1.22 ⇥ 10 4
1 !0 x2 1 ) vy ⇡ 2 ! 0 x
4
=0)x=
8 = 1.6 5
) vy ⇡
* 4. A pulse travelling along a stretched string is described by the following equation: y(x, t) =
b3 b2 + (2x
ut)2
(a) Sketch the graph of y against x for t = 0. (b) What is the speed of the pulse and its direction of travel? (c) The transverse velocity of a given point of the string is defined by vy = @y . Calculate vy as a function of x for the instant t = 0. @t (a) When x = 0, y = b @y 8b3 x = @x [b2 + 4x2 ]2 @y =0 @x )x=0 Turning point at (0, b). x!1 x!
1
1 !0 x2 1 )y⇡ 2 !0 x
)y⇡
note: in this particular case I took b = 2. (b) y(x, t) =
b3 b2 + (x vt)2
In our case we have: b3 y(x, t) = 2 b + (2x b3 = 2 b + 4(x = v=
ut)2 u 2 t) 2
b3 4 b2 4
+ (x
u ms 2
u 2 t) 2
1
(c) @y 4b3 u(2x ut) = 2 @t [b + (2x ut)2 ]2 8b3 ux vy = 2 [b + 4x2 ]2 t=0
vy =
5. Two identical pulses of equal but opposite amplitudes approach each other as they propagate on a string. At t = 0 they are as shown in the figure. Sketch to scale the string, and the velocity profile of string mass elements, at t = 1 sec, t = 1.5 sec, t = 2 sec.
The distance travelled by each wave in a time, t, is x = vt = 10t At t = 0:
At t = 1, x = 1(10)cm, so both waves have moved 10cm:
At t = 1.5, x = 1.5(10)cm, so both waves have moved 15cm:
However, as both waves have the same amplitude they cancel each other out and we get:
At t = 2, x = 2(10)cm, so both waves have moved 20cm:
6. The following two waves in a medium are superposed: y1 = A sin(5x y2 = A sin(4x
10t) 9t)
where x is in meters and t is in seconds. (a) Write an equation for the combined disturbance. (b) What is its group velocity? (c) What is its phase velocity? (a) y = 2A cos [⇡(x k t ⌫)] sin [2⇡(kx ⌫t)] 5 k1 = 2⇡ 4 k2 = 2⇡ 10 ⌫1 = 2⇡ 9 ⌫2 = 2⇡ 4 5 1 k = k1 k2 = = 2⇡ 2⇡ 2⇡ 4 5 + k1 + k2 9 k= = 2⇡ 2⇡ = 2 2 4⇡ 10 9 1 ⌫ = ⌫1 ⌫2 = = 2⇡ 2⇡ 2⇡ 10 9 + ⌫1 + ⌫2 19 ⌫= = 2⇡ 2⇡ = 2 ✓ ✓2 ◆ 4⇡ ✓ ◆◆ ✓✓ ◆ 1 1 9 y = 2A cos ⇡ x t sin 2⇡ x 2⇡ 2⇡ 4⇡ ✓ ◆ ✓ ◆ x t 9x 19t = 2A cos sin 2 2 2 2 (b) vg =
⌫ = k
1 2⇡ 1 2⇡
=1
=
19 9
(c) ⌫ vp = = k
19 4⇡ 9 4⇡
✓
19 4⇡
◆ ◆ t
Vibrations and Waves MP205, Assignment 10 Solutions * 1. Two strings, of tension T and mass densities µ1 and µ2 , are connected together. Consider a traveling wave incident on the boundary. Given the following equations for the transverse displacements in the two strings ✓ ◆ ✓ ◆ x x y1 (x, t) = f1 t + g1 t + v1 v1 ✓ ◆ x y2 (x, t) = f2 t v2 derive the expressions for the ratio of the reflected amplitude to the incident amplitude, and the ratio of the transmitted amplitude to the incident amplitude. Assuming partial reflection and partial ments of the two strings are given by: ✓ y1 (x, t) = f1 t ✓ y2 (x, t) = f2 t
transmission at the junction, the displace◆ ✓ ◆ x x + g1 t + v1 v1 ◆ x v2
At the junction we impose the following conditions to insure that the strings join with equal slopes and tension. y1 (0, t) = y2 (0, t) @y1 @y2 (0, t) = (0, t) @x @x Imposing these conditions implies that we must have: f1 (t) + g1 (t) = f2 (t) 1 0 1 0 1 f1 (t) g1 (t) = f20 (t) v1 v1 v2 We rewrite equation 2 before integrating it. 1 0 1 [f1 (t) g10 (t)] = f20 (t) v1 v2 1 d 0 1 d 0 [f1 (t) g10 (t)] = f (t) v1 dx v2 dx 2 v2 [df10 (t) dg10 (t)] dx = v1 df20 (t)dx Z Z Z 0 0 v2 df1 (t)dx v2 dg1 (t)dx = v1 df20 (t)dx v2 f1 (t)
v2 g1 (t) = v1 f2 (t)
We substitute 1 into this result to find: v2 f1 (t) v2 g1 (t) = v1 f1 (t) + v1 g1 (t) (v2 v1 )f1 (t) = (v1 + v2 )g1 (t) g1 (t) v2 v1 = f1 (t) v2 + v1
(1) (2)
And from 1: g1 (t) = f2 (t)
f1 (t). Filling in gives:
v2 f1 (t)
v2 f2 (t) + v2 f1 (t) = v1 f2 (t) 2v2 f1 (t) = (v1 + v2 )f2 (t) f2 (t) 2v2 = f1 (t) v1 + v2
2. Two strings, of tension T and mass densities µ1 and µ2 , are connected together. Consider a traveling wave incident on the boundary. Find the ratio of the reflected amplitude to the incident amplitude, and the ratio of the transmitted amplitude to the incident amplitude, for the cases µ2 /µ1 = 0, 0.25, 1, 4, 1. v= v1 = v2 =
s s s
T µ T µ1 T µ2
v uT r v1 u µ2 µ1 t = = T v2 µ1 µ2
So we can rewrite our ratios in terms of
µ2 : µ1
v2 v1 g1 (t) = f1 (t) v2 + v1 1 vv12 = 1 + vv12 q µ2 1 µ1 q = 1 + µµ21
For
µ2 µ1
= 0,
q
µ2 µ1
f2 (t) 2v2 = f1 (t) v1 + v2 2 = v1 +1 v2 2 =q µ2 +1 µ1
= 0 so we obtain:
1
q
µ2
µ1 g1 (t) 1 0 q = = =1 f1 (t) 1+0 1 + µµ21
f2 (t) 2 2 =q = =2 µ 2 f1 (t) 0+1 +1 µ1
For
µ2 µ1
= 0.25 = 14 ,
q
µ2 µ1
1 2
=
so we obtain: q
1
µ2
µ1 1 g1 (t) q = = f1 (t) 1+ 1 + µµ21
f2 (t) 2 =q = µ2 f1 (t) +1
1 2
µ1
For
µ2 µ1
= 1,
q
µ2 µ1
1 2 1 2
=
1 2 3 2
=
1 3
2 2 4 = 3 = 3 +1 2
= 1 so we obtain: q
1
µ2
µ1 g1 (t) 1 1 q = = =0 f1 (t) 1+1 1 + µµ21
f2 (t) 2 2 =q = =1 µ 2 f1 (t) 1+1 +1 µ1
For
µ2 µ1
= 4,
q
µ2 µ1
= 2 so we obtain: q
1
µ2
µ1 g1 (t) 1 2 q = = = f1 (t) 1+2 1 + µµ21
1 3
f2 (t) 2 2 2 =q = = µ2 f1 (t) 2+1 3 +1 µ1
For
µ2 µ1
= 1,
q
µ2 µ1
= 1.
The best way to do this is to set 1
q
µ2
q
µ2 µ1
µ1 g1 (t) 1 t q = = = f1 (t) 1+t 1 + µµ21
= t and take the limit as t ! 1: 1 t 1 t
1 = lim + 1 t!1
2 f2 (t) 2 2 =q = = t µ2 f1 (t) t+1 1+ +1 µ1
1 t
= lim
t!1
1 t 1 t
1 0 1 = = 0+1 +1 2 t
1+
1 t
=
1
0 =0 1+0
3. Two strings, of tension T and mass densities µ1 and µ2 , are connected together. Consider a traveling wave incident on the boundary. Show that the energy flux of the reflected wave plus the energy flux of the transmitted wave equals the energy flux of the incident wave. [Hint: The energy flux of a wave (the energy density times the wave speed) is proportional to A2 /v, where A is the amplitude and v is the wave speed.] A2 v A2 =C v
/
where C is a constant
To show: + f2 = f1 (g1 ) (f2 )2 (f1 )2 C +C =C v1 v2 v1 (g1 )2 (f2 )2 (f1 )2 + = v1 v2 v1 2 2 1 (g1 ) 1 (f2 ) 1 + = 2 2 v1 (f1 ) v2 (f1 ) v1 ✓ ◆2 ✓ ◆2 1 g1 1 f2 1 + = v1 f 1 v2 f 1 v1 ✓ ◆2 ✓ ◆2 1 v2 v1 1 2v2 1 + = v1 v2 + v1 v2 v1 + v2 v1 ✓ 2 ◆ ✓ ◆ 2 2 v2 2v2 v1 + v1 4v2 v2 + v1 = v2 2 2 2 v2 + 2v2 v1 + v1 v2 + 2v2 v1 + v12 v2 v22 2v2 v1 + v12 + v1 4v22 = v2 v22 + 2v2 v1 + v12 g1
2
v22
2v2 v1 + v12 + v1 (4v2 ) = v22 + 2v2 v1 + v12 v22 2v2 v1 + v12 + 4v1 v2 = v22 + 2v2 v1 + v12 v22 + 2v2 v1 + v12 = v22 + 2v2 v1 + v12 QED
* 4. A police car, traveling at 60 mi/hr, passes an innocent bystander while sounding its siren, which has a frequency of 2000 Hz. What is the overall change of frequency of the siren as heard by the bystander? Using: ⌫0 ⌫(✓) = u cos ✓ 1 v We have the following values: u = 60mi/hr = 26.82m/s v = 344m/s ⌫0 = 2000Hz 2000 2000 ) ⌫(✓) = = 26.82 cos ✓ 1 0.08 cos ✓ 1 344
To work out the overall change in frequency we need ✓ at the point furthest from the bystander on either side of him. (We’re assuming the police car moves from left to right) As x increases, ↵ decreases, so if we imagine the point furthest from the bystander on the left to be the point where x ! 1 we can take ↵ = 0. As y increases, increases, so if we imagine the point furthest from the bystander on the right to be the point where y ! 1 we can take ↵ = ⇡. 2000 1 0.08 cos ✓ 2000 2000 ⌫(0) = = = 2173.9Hz 1 0.08 cos(0) 1 0.08 2000 2000 ⌫(⇡) = = 1851.86Hz 1 0.08 cos(⇡) 1 + 0.08 ⌫ = 2173.9 1851.86 = 322.04Hz ⌫(✓) =
5. A car, traveling at 65 mi/hr, passes 8m in front of a man while blaring it’s horn continuously, which has a frequency of 1800 Hz. It crashes into a wall 0.2s after passing the man, what is the overall change of frequency of the horn as heard by the bystander? Using: ⌫(✓) =
⌫0 1
u cos ✓ v
We have the following values: u = 65mi/hr = 29.06m/s v = 344m/s ⌫0 = 1800Hz 1800 1800 ) ⌫(✓) = = 29.06 cos ✓ 1 0.084 cos ✓ 1 344
To work out the overall change in frequency we need ✓ at the point furthest from the bystander on either side of him. (We’re assuming the car moves from left to right)
Again, as x increases, ↵ decreases, so if we imagine the point furthest from the bystander on the left to be the point where x ! 1 we can take ↵ = 0. However, unlike last time, we have a definite stopping point for the car. We know when the car it is in front of the man it is 8m in front of him, and we know it stops a distance (29.06)(0.2) = 5.8m to the right of him. Plotting this on a triangle, we need to work out the angle ' and then = ⇡ = .
8 = 1.4 5.8 ' = 0.9 = ⇡ ' = 2.24
tan ' =
1800 1 0.084 cos ✓ 1800 1800 ⌫(0) = = = 1965.07Hz 1 0.084 cos(0) 1 0.084 1800 1800 ⌫(⇡) = = 1714.29Hz 1 0.084 cos(2.24) 1 + 0.05 ⌫ = 1965.07 1714.29 = 250.78Hz ⌫(✓) =
