Vibrations and Waves MP205, Assignment 10 Solutions * 1. Two Two strings, strings, of tension tension T and mass densities µ 1 and µ 2 , are connected together. Consider a traveling wave wave incident incident on the boundary b oundary.. Given the following equations for the transverse displacements in the two strings
y1 (x, t) = f 1 t − y2 (x, t) = f 2 t −
x v1 x v2
+ g1 t +
x v1
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 transmission at the junction, the displacements of the two strings are given by:
y1 (x, t) = f 1 t − y2 (x, t) = f 2 t −
x v1 x v2
+ g1 t +
x v1
At the junction we impose the following conditions to insure that the strings join with equal slopes and tension. y1 (0, (0, t) = y 2 (0, (0, t) ∂y 1 ∂y 2 (0, (0 , t) = (0, (0 , t) ∂x ∂x Imposing these conditions implies that we must have: f 1 (t) + g + g1 (t) = f 2 (t) ∂y 1 ∂y 2 (x, ( x, t) = (x, ( x, t) ∂x ∂x
∂f 1 t
x − v1
∂x Using the substitutions:
∂g 1 t + +
∂x
x v1
(2)
∂f 2 t − =
(1)
x v2
(3)
∂x
(4) x v1 x = t + + u2 = t v1 x u3 = t = t − v2 u1 = t = t −
⇒
⇒
⇒
∂u 1 1 =− ∂x v1 ∂u 2 1 = ∂x v1 ∂u 3 1 =− ∂x v2
(5) (6) (7)
We can rewrite this using the chain rule: (8) 1 ∂f 1 (u1 ) 1 ∂g 1 (u2 ) 1 ∂f 2 (u3 ) + =− v1 ∂u 1 v1 ∂u 2 v2 ∂u 3 At x At x = 0, we can see these substitutions simplify to u 1 = u = u 2 = u 3 = t = t
(9)
−
(10) −
1 df 1 (t) 1 dg1 (t) 1 df 2 (t) + =− v1 dt v1 dt v2 dt 1 1 1 f 1 (t) − g1 (t) = f 2 (t) v1 v1 v2
(11) (12)
We rewrite equation 2 before integrating it. 1 1 f 1 (t) − g1 (t) = f 2 (t) v1 v2 1 d 1 d f 1 (t) − g1 (t) = f (t) v1 dx v2 dx 2 v2 df 1 (t) − dg1 (t) dx = v 1 df 2 (t)dx v2
df 1 (t)dx − v2
dg1 (t)dx = v 1
df 2 (t)dx
v2 f 1 (t) − v2 g1 (t) = v 1 f 2 (t)
We substitute 1 into this result to find: v2 f 1 (t) − v2 g1 (t) = v 1 f 1 (t) + v1 g1 (t) (v2 − v1 )f 1 (t) = (v1 + v2 )g1 (t) g1 (t) v2 − v1 = f 1 (t) v2 + v1 And from 1: g1 (t) = f 2 (t) − f 1 (t). Filling in gives: v2 f 1 (t) − v2 f 2 (t) + v2 f 1 (t) = v 1 f 2 (t) 2v2 f 1 (t) = (v1 + v2 )f 2 (t) f 2 (t) 2v2 = f 1 (t) v1 + v2
2. Two strings, of tension T and mass densities µ 1 and µ 2 , 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, ∞.
v = v1 = v2 = v1 = v2
T µ
T µ1 T µ2 T µ1 T µ2
=
µ2 µ1
µ2 µ1 :
So we can rewrite our ratios in terms of
g1 (t) v2 − v1 = f 1 (t) v2 + v1 1 − vv = 1 + vv 1 2 1 2
µ2 µ1
1− =
µ2 µ1
1+
f 2 (t) 2v2 = f 1 (t) v1 + v2 2 = v v +1 2 = µ µ +1 1 2
2 1
For
µ2 µ1
= 0,
µ2 µ1
= 0 so we obtain:
µ2 µ1
1− g1 (t) = f 1 (t) 1+ f 2 (t) = f 1 (t) For
µ2 µ1
= 0.25 = 14 ,
µ2 µ1
2
µ2 µ1
g1 (t) = f 1 (t) 1+ f 2 (t) = f 1 (t) µ2 µ1
= 1,
µ2 µ1
µ2 µ1 µ2 µ1
2
µ2 µ1
+1
f 2 (t) = f 1 (t) For
= 4,
µ2 µ1
2 =2 0+1
1− = 1+ =
µ2 µ1
g1 (t) = f 1 (t) 1+
=
1 2
1 2 1 2
=
1 2 3 2
=
µ2 µ1
2
µ2 µ1
+1
=
1−1 =0 1+1
=
2 =1 1+1
= 2 so we obtain: 1−
g1 (t) = f 1 (t) 1+ f 2 (t) = f 1 (t)
µ2 µ1 µ2 µ1
2
µ2 µ1
+1
1 3
2 2 4 = 3 = 3 + 1 2
= 1 so we obtain: 1−
µ2 µ1
+1
1−0 =1 1+0
= 12 so we obtain: 1−
For
µ2 µ1
=
=
1−2 1 =− 1+2 3
=
2 2 = 2+1 3
For
µ2 µ1
= ∞,
µ2 µ1
= ∞.
The best way to do this is to set 1−
g1 (t) = f 1 (t) 1+ f 2 (t) = f 1 (t)
µ2 µ1 µ2 µ1
2
µ2 µ1
µ2 µ1
= t and take the limit as t 1
1−t = = 1 + t
1 = lim + 1 t→∞
t −
1
t
2 2 = = t t + 1 1+ +1
1
1 0−1 = = −1 0+1 +1
t −
1
t
2
= lim
1
t→∞
t
:
→ ∞
t
1+
1
t
=
0 =0 1+0
3. Two strings, of tension T and mass densities µ 1 and µ 2 , 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 A 2 /v , where A is the amplitude and v is the wave speed.] A2 Φ∝ v A2 Φ = C where C is a constant v To show: Φg + Φf = Φf 1
2
2
1
(g1 ) (f 2 ) (f 1 )2 C + C = C v1 v2 v1 2 2 (g1 ) (f 2 ) (f 1 )2 + = v1 v2 v1 2 2 1 (g1 ) 1 (f 2 ) 1 + = 2 2 v1 (f 1 ) v2 (f 1 ) v1
v2
2
1 v1
2
2
1 v2 − v1 v1 v2 + v1 2 v2 − 2v2 v1 + v12 + v1 v22 + 2v2 v1 + v12
g1 f 1
2
1 + v2
f 2 f 1
1 + v2
2v2 v1 + v2 4v22 v22 + 2v2 v1 + v12
=
1 v1
=
1 v1
2
= v 2
v2 v22 − 2v2 v1 + v12 + v1 4v22 = v 2 v22 + 2v2 v1 + v12 v22 − 2v2 v1 + v12 + v1 (4v2 ) = v22 + 2v2 v1 + v12 v22 − 2v2 v1 + v12 + 4v1 v2 = v 22 + 2v2 v1 + v12
v22 + 2v2 v1 + v12 = v 22 + 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: ν (θ) =
ν 0 1−
u cos θ v
We have the following values: u = 60mi/hr = 26.82m/s v = 344m/s ν 0 = 2000Hz 2000 2000 = ⇒ ν (θ) = θ 1 − 0.08cos θ 1 − 26.82cos 344
To work out the overall change in frequency we need θ 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, α decreases, so if we imagine the point furthest from the bystander on the left to be the point where x → ∞ we can take α = 0. As y increases, β increases, so if we imagine the point furthest from the bystander on the right to be the point where y → ∞ we can take α = π.
2000 1 − 0.08 cos θ 2000 2000 ν (0) = = = 2173.9Hz 1 − 0.08 cos(0) 1 − 0.08 2000 2000 = 1851.86Hz ν (π) = 1 − 0.08 cos(π) 1 + 0.08 ∆ν = 2173.9 − 1851.86 = 322.04Hz ν (θ) =
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: ν (θ) = We have the following values:
ν 0 1−
u cos θ v
u = 65mi/hr = 29.06m/s v = 344m/s ν 0 = 1800Hz 1800 1800 = ⇒ ν (θ) = θ 1 − 0.084 cos θ 1 − 29.06cos 344
To work out the overall change in frequency we need θ 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, α decreases, so if we imagine the point furthest from the bystander on the left to be the point where x → ∞ we can take α = 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 ϕ and then β = π − φ.
8 = 1.4 5.8 ϕ = 0.9
tan ϕ =
β = π − ϕ = 2.24 1800 1 − 0.084 cos θ 1800 1800 ν (0) = = = 1965.07Hz 1 − 0.084 cos(0) 1 − 0.084 1800 1800 ν (2.24) = = 1714.29Hz 1 − 0.084 cos(2.24) 1 + 0.05 ∆ν = 1965.07 − 1714.29 = 250.78Hz ν (θ) =
