Chapter # 16
1. Sol.
Sound Waves
SOLVED EXAMPLES
A wave of wavelength 0.40 cm is produced in air and it travels at a speed of 300 m/s. Will it be audible ? From the relation v = , the frequency of the wave is
300 m / s = = 75000 Hz. 0.40 10 2 m This is much above the audible range. It is an ultrasonic wave and will not be audible to humans, but it will be audible to bats. =
2.
Sol.
A sound wave of wavelength to bats 40 cm travels in air. If the difference between the maximum and minimum pressures at a given point is 2.0 × 10–3 N/m2, find the amplitude of vibration of the particles of the medium. The bulk modulus of air is 1.4 × 105 N/m2. The pressure amplitude is
2.0 10 3 N / m 2 = 10–3 n/M2. 2 The displacement amplitude s0 is given by p0 = B k s0 p0 =
or,
p0 p0 s0 = B k = 2 B =
10 3 N / m 2 ( 40 10 2 m)
2 3.14 14 10 6 N / m 2 = 6.6 Å
3.
Sol.
The pressure amplitude in a sound wave from a radio receiver is 2.0 × 10–3 N/m2 and the intensity at a point is 10–6 W/m2. If by turning the “Volume” knob the pressure amplitude is increased to 3 × 10–3 N/m2, evaluate the intensity. The intensity is proportional to the square of the pressure amplitude. Thus,
or,
p0 = p 0
2
2
p0 = p I = 0
2 3 2.0
× 10–6 W/m2
= 2.25 × 10–16 W/m2. 4.
If the intensity is increased by a factor of 20, by how many decibels is the level increased.
Sol.
Let the initial intensity by I and the second level be 1. When the intensity is increased to 20 the level increases to 2. Then 1 = 10 log (I / I0) and 2 = 10 log (20 / 0) Thus, 2 – 1 = 10 log (20 /) = 10 log 20 = 13 dB.
5.
Two sound waves, originating from the same source, travel along different paths in air and then meet at a point. If the source vibrates at a frequency of 1.0 kHz and one path is 83 cm longer than the other, what will be the natural of interference ? The speed of sound in air is 332 m/s.
Sol.
The wavelength of sound wave is =
u v
332 m / s
=
1.0 10 3 Hz
= 0.332 m.
The phase difference between the waves arriving at the point of observation is
manishkumarphysics.in
Page # 1
Chapter # 16
Sound Waves =
0.83 m 2 x = 2 × 0.332 m = 2 × 2.5 = 5.
As this is an odd multiple of , the waves interfere destructively. 6.
An air column is constructed by fitting a movable piston in a long cylindrical tube. Longitudinal waves are sent in the tube by a tuning fork of frequency 416 Hz. How far from the open end should the piston be so that the air column in the tube may vibrate in its first overtone? Speed of sound in air is 333 m/s.
Sol.
The piston provides the closed end of the column and an antinode of pressure is formed there. At the open end, a pressure node is formed. In the first overtone there is one more node and an antinode in the column as shown in figure. The length of the tube should then be 3/4.
333 m / s u = = 0.800 m. 416 s 1 v Thus, the length of the tube is
The wavelength is =
3 3 0.800m = = 60.0 cm. 4 4 7.
A tuning fork A of frequency 384 Hz gives 6 beats in 2 seconds when sounded with another tuning fork B. What could be the frequency of B?
Sol.
The frequency of beats is | v1 – v2 |, which is 3 Hz according to the problem. The frequency of the tuning fork B is either 3 Hz more or 3 Hz less than the frequency of A. Thus, it could be either 381 Hz or 387 Hz.
8.
A sound detector is placed on a railway platform. A train, approaching the platform at a speed of 36 km/h, sounds its whistle. The detector detects 12.0 kHz as the most dominant frequency in the whistle. If the train stops at the platform and sounds the whistle, what would be the most dominant frequency detected ? The speed of sound in air is 340 m/s.
Sol.
Here the observer (detector) is at rest with respect to the medium (air). Suppose the dominant frequency as emitted by the train is v0. When the train is at rest at the platform, the detector will detect the dominant frequency as v0. When this same train was approaching the observer, the frequency detected was,
v v = v u v0 s
or,
v0 =
us v us v. v = 1 v v
The speed of the source is us = 36 km/h =
Thus,
36 10 3 m = 10 m/s. 3600 s
10 × 12.0 kHz v0 = 1 340
QUESTIONS
= 11.6 kHz.
FOR
SHORT
ANSWER
1.
If you are walking on the moon, can you hear the sound of stones cracking behind you? Can you hear the sound of your own footsteps?
2.
Can you hear your own wards if you are standing in a perfect vacuum? Can you hear your friend in the same conditions?
3.
A vertical rod is hit at one end. What kind of wave propagates in the rod is (a) the hit is made vertically (b) the hit is made horizontally? manishkumarphysics.in
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Chapter # 16
Sound Waves
4.
Two loudspeakers are arranged facing each other at some distance. Will a person standing behind one of the loudspeakers clearly hear the sound of the other loudspeaker or the clarity will be seriously damaged because of the ‘collision’ of the two sounds in between ?
5.
The voice of a person, who have inhaled helium, has a remarkably high pitch. Explain on the basis of resonant vibration of vocal cord filled with air and with helium.
6.
Draw a diagram to show the standing pressure wave and standing displacement wave for the 3rd overtone mode of vibration of an open organ pipe.
7.
Two tuning forks vibrate with the same amplitude but the frequency of the first is double the frequency of the second. Which fork produces more intense sound in air?
8.
In discussing Doppler effect, we use the word “apparent frequency”. Does it mean that the frequency of the sound is still that of the source and it is some physiological phenomenon in the listener’s ear that gives rise to Doppler effect? Think for the observer approaching the source and for the source approaching the observer.
Objective - I 1.
Consider the following statements about sound passing through a gas. (A*) the pressure of the gas at a point oscillates in time (B) A is correct but B is wrong (C) B is correct but A is wrong (D) Both A and B are wrong
fdlh xSl ls xqtjus okyh /ofu ds fy;s fuEu dFkuksa ij fopkj dhft;saA (A*) fdlh fcUnq ij xSl dk nkc] le; ds lkFk nksy u djrk gSA (a) A o B nksu ksa lR; gSA (b) A lRl gS fdUrq B vlR; gS (c) B lR; gS fdUrq A vlR; gSA (d) A o B nksu ksa gh vlR; gSA 2.
When we clap our hands, the sound produced is best described by [HCV_chp 16_obj 1_2] (A) p = p0 sin (kx – t) (B) p = p0 sin kx cos t (C) p = p0 cos kx sin t (D*) p = ponsin (knx – nt) Here p denotes the change in pressure from the equilibrium value.
tc ge gkFkksa ls rkyh ctkrs gS] mRiUu /ofu dk lokZf/kd lg fu:i.k fuEu ds }kjk gksrk gSA (A) p = p0 sin (kx – t) (C) p = p0 cos kx sin t ;gk¡ p lkE;koLFkk ls nkc ds 3.
(B) p = p0 sin kx cos t (D*) p = ponsin (knx – nt)
eku esa ifjorZu dk O;Dr djrk gSA
The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air
ty dk vk;ru izR;kLFkrk xq.kkad rFkk ?kuRo] ok;q ls vf/kd gSA ek=k bruh lwpuk ds vk/kkj ij ge dg ldrs gSA ek=k bruh lwpuk ds vk/kkj ij] ge dg ldrs gS fd ok;q esa /ofu dk osx & (A) is larger than its value in water (C) is equal to its value in water (A) ty esa blds eku ls vf/kd gS (C) ty esa blds eku ds cjkcj gS 4.
(B) is smaller than its value in water (D*) cannot be compared with its value in water (B) ty esa blds eku ls de gSA (D*) ty esa blds eku ls rqy uk ugha dh tk ldrh
gSA
A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change ? (A) displacement amplitude (B) Frequency (C*) Wavelength (D) time period
LofL=k f}Hkqt ok;q esa /ofu rjaxs izlkfjr djrk gSA ;fn ok;q dk rki c<+ tk;s] fuEu esa ls dkSulh jkf'k ifjofrZr gksxh& (A) foLFkkiu vk;ke (B) vko`fÙk (C*) rjax nS/;Z (D) vkorZdky 5.
When sound wave is refracted from air to water, which of the following will remain unchanged? (A) wave number (B) wavelength (C) wave velocity (D*) frequency manishkumarphysics.in
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Chapter # 16
Sound Waves
tc /ofu rjaxs ok;q ls ty esa viofrZr gksrh gS] fuEu esa ls D;k vifjofrZr jgrk gS& (A) rjax la[;k (B) rja x nS / ;Z (C) rjax osx (D*) vko`fÙk 6.
The speed of sound in a medium depends on (A) the elastic property but not on the inertia property (B) the inertia property but not on the elastic property (C*) the elastic property as well as the inertia property (D) neither the elastic property nor the inertia property
fdlh ek/;e esa /ofu dk osx fuHkZj djrk gS& (A) izR;kLFk xq.k ij fdUrq tM+Roh; xq.k ij ugh (B) tM+Roh; xq.k ij fdUrq izR;kLFk xq.k ij ugh (C*) izR;kLFk xq.k ds lkFk tMRoh; xq.k ij Hkh (D) u rks izR;kLFk xq.k ij u gh tM+Roh; xq.k ij 7.
Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P1 and that by the second wave be P2. Then
,d gh ek/;e esa nks /ofu rjaxsa ,d gh fn’kk esa xeu dj jgh gSA rjaxksa ds nkc&vk;ke leku gS fdUrq izFke rjax dh rjaxnS/;Z nwljh dh nqxuh gSA ekuk fd fdlh vuqizLFk dkV ls izFke rjax }kjk LFkkukarfjr 'kfä P1 gSA ,oa nwljh rjax }kjk P2 gS] rks & (A*) P1 = P2 8.
(B) P1 = 4P2
(C) P2 = 2P1
(D) P2 = 4P1
When two waves with same frequency and constant phase difference interfere,
tc leku vko`fÙk rFkk fu;r dykUrj okyh nks rjaxsa O;frdj.k djrh gS] rks & (A) there is a gain of energy ÅtkZ izkIr gksrh gSA (B) there is a loss of energy ÅtkZ dh gkfu gksrh gSA (C) the energy is redistributed and the distribution changes with time
ÅtkZ dk iqufoZrj.k gksrk gS rFkk le; ds lkFk forj.k ifjofrZr gksrk gSA (D*) the energy is redistributed and the distribution remains constant in time
ÅtkZ dk iqufoZrj.k gksrk gS rFkk le; ds lkFk forj.k fu;r jgrk gSA 9.
An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximum (A) at the two ends (B*) at the middle of the pipe (C) at distance L/4 inside the ends (D) at distance L/8 inside the ends L yEckbZ okyk ,d [kqy k vkxZu ikbi bldh ewy vko`fÙk ls dEiUu dj jgk gSA nkc esa ifjorZu vf/kdre gksxk& (A) nksu ksa fljksa ij (B*) ikbi ds e/; esa (C) ,d fljs ls vUnj dh vksj L/4 nwjh ij (D) ,d fljs ls vUnj dh vksj L/8 nwjh ij
10.
An organ pipe, open at both ends, contains (A*) longitudinal stationery waves (B) longitudinal travelling waves (C) transverse stationary waves (D) transverse travelling waves
nksuksa fljksa ij [kqys gq, fdlh vkxZu ikbi esa& (A*) vuqnS/;Z vizxkeh rjaxs gksrh gSA (C) vuqizL Fk vizxkeh rjaxs gksrh gSA 11.
(B) vuqnS/;Z (D) vuqizL Fk
izxkeh rjaxs gksrh gSA izxkeh rjaxs gksrh gSA
A cylindrical tube, open at both ends, has a fundamental frequency . The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is nksuksa fljksa ij [kqyh gqbZ ,d csyukdkj uyh dh ewy vko`fÙk gSA uyh dks ikuh esa bl izdkj Mqck;k x;k gSA fd bldh
vk/kh yEckbZ ikuh ds vUnj gSA u;h ewy vko`fÙk gksxh& (A) /4 12.
(B) /2
(C*)
(D) 2
The phenomenon of beats can take place (A) for longitudinal wave only (C*) for both longitudinal and transverse waves
(B) for transverse wave only (D) for sound waves only.
foLian dh ?kVuk gks ldrh gSA (A) dso y vuqnS/;Z rjaxks esa (C*) vuqnS/;Z o vuqnS/;Z nksu ksa gh rjaxks esa
(B) dso y (D) dso y
manishkumarphysics.in
vuqizLFk rjaxks esa /ofu rjaxks esa Page # 4
Chapter # 16 13.
Sound Waves
A tuning fork of frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is ,d Lojekih ds rkj ds lkFk 512 gVZ~t vko`fÙk okyk Lofj=k f}&Hkqt dfEir djokus ij 6 foLian izfr lsd.M+ lquk;h
nsrs gSA ;fn rkj dk ruko FkksM+k c<+k fn;k tk;s rks foLian vko`fÙk de gks tkrh gSA rkj dh ewy vko`fÙk gS& 14.
15.
(A*) 506 Hz (B) 512 Hz (C) 518 Hz (D) 524 Hz The engine of a train sounds a whistle at frequency . The frequency heard by a passenger is ,d Vsªu dk batu vko`fÙk dh lhVh ctkrk gS] ,d ;k=kh }kjk lquh x;h vko`fÙk gksxh& (A) > (B) < (C) =(1 / (D*) = The change in frequency due to Doppler effect does not depend on (A) the speed of the source (B) the speed of the observer (C) the frequency of the source (D*) separation between the source and the observer
MkWIyj izHkko ds dkj.k vko`fÙk esa ifjorZu fuHkZj ugh djrk gS& (A) L=kksr dh pky ij (B) izs{k.k dh pky ij (C) L=kksr dh vko`fÙk ij (D*) L=kksr rFkk izs{k.k ds e/; dh nwjh ij 16.
A small source of sound moves on a circle as shown in fig. and an observer is sitting at O. Let at be the frequencies heard when the source is at A, B, and C respectively. /ofu dk ,d NksVk L=kksr iznf'kZr fp=kkuqlkj o`Ùk ij xfr dj jgh gS] rFkk izs{k.k O ij fLFkr gSA ekukfd lquh x;h vko`fÙk;k¡ gS] tc L=kksr Øe'k% A, B, o C ij gS&
(A) 1 > 2 > 3
(B) 1 = 2 > 3
(C) 2 > 3 > 1
(D*)
1 > 3 > 2
Objective - II 1.
When you speak to your friend, which of the following parameters have a unique value in the sound produced? (A) Frequency (B) Wavelength (C) Amplitude (D*) Wave velocity
tc vius fe=k ls ckr djrs gS] mRiUu /ofu esa fuEu ls fdu jkf'k;ksa ds ,dy eku gksrs gS& (A) vko`fÙk (B) rja x nS / ;Z (C) vk;ke (D*) rjax osx 2.
An electrically maintained tuning fork vibrates with constant frequency and constant amplitude. If the temperature of the surrounding air increases but pressure remains constant, the sound produced will have (A*) large wavelength (B) larger frequency (C*) larger velocity (D) larger time period
,d fo/kqr iksf"kr Lofj=k f}&Hkqt fu;r vko`fÙk ,oa fu;r vk;ke ls dEiUu dj jgk gSA ;fn vkl&ikl dh ok;q dk rki ifjofrZr gks tk;s fdUrq nkc vifjofrZr jgs] rks mRiUu /ofu& (A*) dh rjaxnS/;Z vf/kd gksxhA (B) dh vko`fÙk vf/kd gksxhA (C*) dk osx vf/kd gksxkA (D) dk vkorZdky vf/kd gksxkA 3.
4.
The fundamental frequency of a vibrating organ pipe is 200 Hz. (A) The first overtone is 400 Hz. (B*) The first overtone may be 400 Hz. (C*) The first overtone may be 600 Hz (D*) 600 Hz is an overtone dEiUu dj jgs vkxZu ikbi dh ewy vko`fÙk 200 gV~Zt gS& (A) izFke lUukna 400 gV~Zt dk gksxkA (B*) izFke lUukn 400 gV~Zt dk gks ldrk (C*) izFke lUukn 600 gV~Zt dk gks ldrk gSA (D*) dksbZ lUukn 600 gV~Zt dk gksxkA
gSA
A source of sound moves towards an observer (A) The frequency of the source is increased (B) The velocity of sound in the medium is increased (C*) The wavelength of sound in the medium towards the observer is decreased (D) The amplitude of vibration of the particles is increased
,d /ofu L=kksr] izs{kd dh vksj xfr'khy gS& (A) L=kksr dh vko`fÙk c<+ tk;sxhA manishkumarphysics.in
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Chapter # 16
Sound Waves
(B) ek/;e esa /ofu dk osx c<+ tk;sxkA (C*) isz{k.k dh vksj ek/;e esa /ofu dh rjaxnS/;Z (D) d.kksa ds dEiUu dk vk;ke c<+ tk;sxkA 5.
de gks tk;sxhA
A listener is at rest with respect to the source of sound. A wind starts blowing along the line joining the source and the observer. Which of the following quantities do not change? (A*) Frequency (B) Velocity of sound (C) Wavelength (D*) Time period
,d Jksrk] /ofu L=kksr ds lkis{k fLFkj gSA L=kksr rFkk isz{kd dks feykus okyh js[kk ds vuqfn'k ok;q izokfgr gksus yxrh gSA fuEu esa ls dkSulh jkf'k ifjofrZr ugh gksxh& (A*) vko`fÙk (B) /ofu dk osx (C) rjax nS /;Z (D*) vkorZdky
WORKED OUT EXAMPLES
1.
Sol.
An ultrasound signal of frequency 50 kHz is sent vertically into sea water. The signal gets reflected from the ocean bed and returns to the surface 0.80 s after it was emitted. The speed of sound in sea water is 1500 m/ s. (a) Find the depth of the sea. (b) What is the wavelength of this signal in water ? (a) Let the depth of the sea be d. The total distance travelled by the signal is 2d. By the question, 2d = (1500 m/s) (0.8 s) = 1200 m or, d = 600 m. (b) Using the equation u = v, =
1500 m / s u = = 3.0 cm. 50 10 3 s 1 v
2.
An aeroplane is going towards east at a speed of 510 km/h at a height of 2000 m. At a certain instant, the sound of the plane heard by a ground observer appears to come from a point vertically about him. Where is the plane at this instant ? Speed of sound in air = 340 m/s.
Sol.
The sound reaching the ground observer P, was emitted by the plane when it was at the point Q vertically above his head. The time taken by the sound to reach the observer is
2000 m 100 t = 340 m / s = s. 17 The distance moved by the plane during this period is 100 30 10 5 s d = (510 km/h = m = 833 m. 17 3600 Thus, the plane will be 833 m ahead of the observer on its line of motion when he hears the sound coming vertically to him.
3.
Sol.
The equation of a sound wave in air is given by p = (0.01 N/m2) sin [(1000 s–1) t – (3.0 m–1) x ] (a) Find the frequency, wavelength and the speed of sound wave in air. (b) If the equilibrium pressure of air is 1.0 × 105 N/m2, what are the maximum and minimum pressures at a point as the wave passes through that point? (a) Comparing with the standard form of a travelling wave p = p0 sin [w (t – x/v)] we see that w = 1000 s–1. The frequency is 1000 = Hz = 160 Hz. 2 2 Also from the same comparison, w/v = 3.0 m–1.
v=
or,
v=
3.0 m 1
=
1000 s 1 3.0 m 1
= 330 m/s. manishkumarphysics.in
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Chapter # 16
Sound Waves
330 m / s u = 160 Hz = 2.1 m. v (b) The pressure amplitude is p0 = 0.01 N/m2. Hence, the maximum and minimum pressures at a point in the wave motion will be (1.01 × 105 ± 0.01) N/m2. The wavelength is =
4. Sol.
A sound wave of frequency 10 kHz is travelling in air with a speed of 340 m/s. Find the minimum separation between two points where the phase difference is 60°. The wavelength of the wave is =
340 m / s u = = 3.4 cm 10 10 3 s 1 v
2 2 = cm–1. 3 .4 The phase of the wave is (kx – t). At any given instant, the phase difference between two points at a separation d is kd. If this phase difference is 60° i.e., /3 radian ;
The wave number is k =
2 cm 1 d = 3 3. 4
5. Sol.
or
3 .4 cm = 0.57 cm. 6
d=
On a winter day sound travels 336 metres in one second. Find the atmospheric temperature. Speed of sound at 0°C = 332 m/s. The speed of sound is proportional to the square root of the absolute temperature. The speed of sound at 0°C or 273 K is 332 m/s. If the atmospheric temperature is T,
336 m / s 332 m / s = 336
T 273 K
2
or,
T = 332 × 273 K = 280 K
or,
t = 7°C.
6.
The constant for oxygen as well as for hydrogen is 1.40. If the second of sound in oxygen is 470 m/sm, what will be the speed in hydrogen at the same temperature and pressure?
Sol.
The speed of sound in a gas is given by u =
P . At STP, 22.4 litres of oxygen has a mass of 32 g whereas
the same volume of hydrogen has a mass of 2 g. Thus, the density of oxygen is 16 times the density of hydrogen at the same temperature and pressure. As is same for both the gases,
u (hydrogen ) u (oxygen) = or,
7.
Sol.
(oxygen ) (hydrogen )
v (hydrogen) = 4u (oxygen) = 4 × 470 m/s = 1880 m/s.
A microphone of cross-sectional area 0.80 cm2 is placed in front of a small speaker emitting 3.0 W of sound output. If the distance between the microphone and the speaker is 2.0 m, how much energy falls on the microphone in 5.0 s ? The energy emitted by the speaker in one second is 3.0 J. Let us consider a sphere of radius 2.0 m centred at the speaker. The energy 3.0 J falls normally on the total surface of this sphere in one second. The energy falling on the area 0.8 cm2 of the microphone in one second =
0.8 cm2 4 (2.0 m)2
× 30 J = 48 × 10–6J.
The energy falling on the microphone in 5.0 is 4.8 × 10–6 J × 5 = 24 µJ. 8.
Sol.
Find the amplitude of vibration of the particles of air through which a sound wave of intensity 2.0 × 10–6 W/m2 and frequency 1.0 kHz is passing. Density of air = 1.2 kg/m3 and speed of sound in air = 330 m/s. The relation between the intensity of sound and the displacement amplitude is manishkumarphysics.in Page # 7
Chapter # 16
Sound Waves = 22 s02v20v
or,
or, 9. Sol.
2
s0 =
=
2 2 v 2 0 v = 2.53 × 10–16 m2 s0 = 1.6 × 10–8 m.
2.0 10 6 W / m 2 2 2 (1.0 10 6 s 2 ) (1.2 kg / m3 ) (330 m / s)
The sound level at a point is increased by 30 dB. By what factor is the pressure amplitude increased ? The sound level in dB is
= 10 lg10 . 0 If 1 and 2 are the sound levels and 1 and 2 are the intensities in the two cases,
2 1 2 – 1 = 10 log10 log10 0 0 or,
2 30 = 10 log10 0
2 2 3 or, 1 = 10 . 1 As the intensity is proportional to the square of the pressure amplitude, or,
we have
2 0 = 10 log10
p2 p1 =
2 1 =
1000 32.
10.
Figure shows a tube structure in which a sound signal is sent from one end and is received at the other end. The semicircular part has a radius of 20.0 cm. The frequency of the sound source can be varied electronically between 1000 and 4000 Hz. Find the frequencies at which maximum of intensity are detected. The speed of sound in air = 340 m/s.
Sol.
The sound wave bifurcates at the junction of the straight and the semicircular parts. The wave through the straight part travels a distance 1 = 2 × 20 cm and the wave through the curved part travels a distance 2 = 20 cm = 62.8 cm before they meet again and travel to the receiver. The path difference between the two waves received is, therefore. = 2 – 1 = 62.8 cm – 40 cm = 22.8 cm = 0.228 m. The wavelength of either wave is
340 m / s = . For constructive interference, = n, where n is an v v
integer.
340 m / s v
or,
0.228 m = n
or,
340 m / s v = n 0.228 m = n 1491.2 Hz = n 1490 Hz.
Thus, the frequencies with in the specified range which cause maximum of intensity are 1490 Hz and 2980 Hz. 11.
A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance of 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound. Speed of sound in air = 360 m/s.
Sol.
The situation is shown in figure. Suppose the detector is placed at a distance of x meter from the source. The direct wave received from the source travels a distance of x meter. The wave reaching the detector after reflection from the wall has travelled a distance of 2(2)2 + x2/4]1/2 metre. The path difference between the two waves is manishkumarphysics.in
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Chapter # 16
Sound Waves 1/ 2 4 2 x x metre. = 2 (2) 4
Constructive interference will take place when = , 2, ...... The minimum distance x for a maximum corresponds to = ..........(i) The wavelength is =
360 m / s = = 2 m. 180 s 1 v
2 x2 Thus, by (i) 2 (2) 4
or,
x2 4 4
1/ 2
–x=2
1/ 2
=1+
x 2
x2 x2 =1+ +x 4 4 or, 3 = x. Thus, the detector should be placed at a distance of 3 m from the source. Note that there is no abrupt phase change. or,
4+
12.
A tuning fork vibrates at 264 Hz. Find the length of the shortest closed organ pipe that will resonate with the tuning fork. Speed of sound in air is 350 m/s.
Sol.
The resonant frequency of a closed organ pipe of length is
nv , where n is a positive odd integer and v is the 4 speed of sound in air. To resonate with the given tuning fork, nv = 264 s–1 4 or,
=
n 350 m / s 4 264 s 1
For to be minimum, n = 1 so that
350
min = 4 264 m = 33 cm. 13.
Sol.
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, what is the length of the closed pipe?
v The fundamental frequency of a closed organ pipe is 4 . For an open pipe, the fundamental frequency is 1 v 2v v and the first overtone is = 2 2 2 2 2 . Here 1 is the length of the closed pipe and l2 = 60 cm is the length of the open pipe. We have, v v = 4 1 60 cm l1 =
14.
1 × 60 cm = 15 cm. 4
A tuning fork vibrating at frequency 800 Hz produces resonance in a resonance column tube. The upper end is open and the lower end is closed by the water surface which can be varied. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. Calculate the speed of sound in air from these data. manishkumarphysics.in
Page # 9
Chapter # 16 Sol.
Sound Waves
nv , where n is a positive odd integer. If the 4 tuning fork has a frequency v and 1, 2, 3 are the successive lengths of the tube in resonance with it, we have For the tube open at one end, the resonance frequencies are
n 4 1 = v
Þ
(n 2 ) =v 4 2
Þ
(n 4) =v 4 3
2 = . 4v 2v By the question, 3 – 2 = (52.75 – 31.25) cm = 21.50 cm and 2 – 1 = (31.25 – 9.75) cm 21.50 cm giving
3 – 2 = 2 – 1 =
thus,
= 21.50 2v = 2v × 21.50 cm = 2 × 21.50 cm = 344 m/s.
or, 15.
Sol.
A certain organ piperesonates in its fundamental mode at a frequency of 500 Hz in air. What will be the fundamental frequency if the air is replaced by hydrogen at the same temperature ? The density of air is 1.20 kg/m3 and that of hydrogen is 0.089 km/m3. Suppose the speed of sound in hydrogen is h and that in air is a. The fundamental frequency of an organ pipe is proportional to the speed of sound in the gas contained in it. If the fundamental frequency with hydrogen in the tube is v, we have
h v = 500 Hz a = or, 16.
Sol.
An aluminium rod having a length of 90.0 cm is clamped at its middle point and is set into longitudinal vibrations by stroking its with a rosined cloths. Assume that the of vibrates in its fundamental mode of vibration. The density of aluminium is 2600 kg/m3 and its Young’s modulus is 7.80 × 1010 N/m2. Find (a) the speed of sound in aluminium, (b) the wavelength of sound waves produced in the rod, (c) the frequency of the sound produced and (d) the wavelength of the sound produced in air. Take the speed of sound in air to be 340 m/s. (a) The speed of sound in aluminium is
(b)
(c)
(d)
7.80 1010 N / m 2 2600 kg / m 3
5480 = 180 cm = 3050 Hz.
The wavelength of sound in air is =
18.
Y =
= 5480 m/s. Since the rod is clamped at the middle, the middle point is a pressure antinode. The free ends of the rod are the nodes. As the rod vibrates in its fundamental mode, there are no other nodes or antinodes. The length of the rod, which is also the distance between the successive nodes, is, therefore, equal to half the wavelength. Thus, the wavelength ofsound in the aluminium rod is = 2 = 180 cm The frequency of the sound produced which is also equal to the frequency of vibration of the rod is v=
Sol.
1 .2 = 3.67 0.089
v = 3.67 × 500 Hz 1840 Hz.
=
17.
a h =
340 m / s = 3050 Hz = 11.1 cm.
The string of a violin emits a note of 440 Hz at its correct tension. The string is bit taut and produces 4 beats per second with a tuning fork of frequency 440 Hz. Find the frequency of the note emitted by this taut string. The frequency of vibration of a string increases with increase in the tension. Thus, the note emitted by the string will be a little more than 440 Hz. As it produces 4 beats per second with the 440 Hz tuning fork, the frequency will be 444 Hz. A siren is fitted on a car going towards a vertical wall at a speed of 36 km/h. A person standing on the ground, behind the car, listens to the siren sound coming directly from the source as well as that coming after manishkumarphysics.in
Page # 10
Chapter # 16 Sound Waves reflection from the wall. Calculate the apparent frequency of the wave(a) coming directly from the siren to the person and (b) coming after reflection. Take the speed of sound to be 340 m/s.
Sol. The speed of the car is 36 km/h = 10 m/s. (a) Here the observer is at rest with respect to the medium and the source is going away from the observer. The apparent frequency heard by the observer is, therefore,
v = u v s
=
340 × 500 Hz = 486 Hz. 340 10
(b) The frequency received by the wall is
340 v= u v = × 500 Hz = 515 Hz. 340 10 s The wall reflects this sound without changing the frequency. Thus, the frequency of the reflected wave as heard by the ground observer is 515 Hz. 19.
Two trains are moving towards each other at speeds of 72 km/h and 54 km/h relative to the ground. The first train sounds a whistle of frequency 600 Hz. Find the frequency of the whistle as heard by a passenger in the second train. (a) before the trains meet and (b) after the trains have crossed each other. The speed of sound in air is 340 m/s.
Sol.
The speed of the first train = 72 km/h = 20 m/s and that of the second = 54 km/h = 15 m/s. (a) Here both the source and the observer move with respect to the medium. Before the trains meet, the source is going towards the observer and the observer is also going towards the source. The apparent frequency heard by the observer will be
u0 v = u v s
=
340 15 × 600 Hz = 666 Hz. 340 20
(b) After the trains have crossed each other, the source goes away from the observer and the observer goes away from the source. The frequency heard by the observer is, therefore,
u0 v = u v s
=
340 15 × 600 Hz = 542 Hz. 340 20
20.
A person going away from a factory on his scooter at a speed of 36 km/h listens to the siren of the factory. If the main frequency of the siren is 600 Hz and a wind is blowing along the direction of the scooter at 36 km/ h, find the main frequency as heard by the person.
Sol.
The speed of sound in still air is 340 m/s. Let us work from the frame of reference of the air. As both the observer and the wind are moving at the same speed along the same direction with respect to the ground, the observer is at rest with respect to the medium. The source (the siren) is moving with respect to the wind at a speed of 36 km/h i.e., 10 m/s. As the source is going away from the observer who is at rest with respect to the medium, the frequency heard is
340 v = u v = × 600 Hz = 583 Hz. 340 10 s 21.
A source and a detector move away from each other, each with a speed of 10 m/s with respect to the ground with no wind. If the detector detects a frequency 1950 Hz of the sound coming from the source, what is the original frequency of the source? Speed of sound in air = 340 m/s.
Sol.
If the original frequency of the source is v, the apparent frequency heard by the observer is
u0 v = u v,, s where u0 is the speed of the observer going away from the source and us is the speed of the source going away from the observer. By the question,
manishkumarphysics.in
Page # 11
Chapter # 16
Sound Waves 1950 Hz =
340 10 v 340 10
v= 22.
35 × 1950 Hz = 2070 Hz. 33
The driver of a car approaching a vertical wall notices that the frequency of his car’s horn changes from 440 Hz to 480 Hz when it gets reflected from the wall. Find the speed of the car if that of the sound is 330 m/s.
Sol. Suppose, the car is going towards the wall at a speed u. The wall is stationary with respect to the air and the horn is going towards it. If the frequency of the horn is v, that received by the wall is v.. u The wall reflects this sound without changing the frequency. Thus, the wall becomes the source of frequency v and the car-driver is the listener. The wall (which acts as the source now) is at rest with respect to the air and the car (which is the observer now) is going towards the wall at speed u. The frequency heard by the cardriver for this reflected wave is, therefore,
v =
v =
u v
=
u . v u
=
u v.. u
Putting the values, 480 =
23.
u 440 u
or,
u 48 = u 44
or,
u=
or,
u 4 = 92
4 × 330 m/s = 14.3 m/s 52 km/h 92
A train approaching a railway crossing at a speed of 120 km/h sounds a short whistle at frequency 640 Hz when it is 300 m away from the crossing. The speed of sound in air is 340 m/s. What will be the frequency heard by a person standing on a road perpendicular to the track through the crossing at a distance of 400 m from the crossing ?
Sol.
The observer A is at rest with respect to the air and the source is travelling at a velocity of 120 km/h i.e., 100 m/s. As is clear from the figure, the person receives the sound of the whistle in a direction BA making 3 an angle with the track where cos = 300/500 = 3/5. The component of the velocity of the source (i.e., of 100 3 × m/s = 20 m/s. As the source is approaching the person with this 3 5 component, the frequency heard by the observer is
the train) along this solution AB is
v =
v 340 v= × 640 Hz = 680 Hz. u cos 340 20
manishkumarphysics.in
Page # 12
Chapter # 16
1.
Sound Waves
EXERCISE
A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound to the blow twice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sound in various substances. LVhy dh 1.00 eh- yEch ,d uyh dks ,d fljs ij Bksdk tkrk gSA ,d O;fDr tks uyh ds nwljs fljs ij dku yxkdj
[kM+k gqvk gSA pksV dh /ofu nks ckj /ofu lqurk gS] ,d uyh ds inkFkZ ds vUnj gksdj vkus okyh rFkk nwljh uyh dh ok;q ls gksdj vkus okyhA nksuksa /ofu;kas ds lqukbZ nsus esa le;karjky Kkr dhft;sA Ans. 2.75 ms 2.
At a prayer meeting, the disciples sing JAI-RAM, JAI-RAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What minimum time interval can be kept between one JAI-RAM and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is 320 m/s. ,d izkFkZuk lHkk esa] HkDr x.k t;&jke]t;&jke xk jgs gSA ykmM+ Lihdj ls izof/kZr /ofu] lHkk ls 80 eh- nwjh ij fLFkr
Hkou ls ijkofrZr gksdj iqu% ykSVrh gSA ,d t;&jke rFkk vxys t;&jke esa fdruk le;kUrjky j[kuk pkfg;s fd lHkk esa cSBs gq, Jksrk dks vuqxwat (echo) ck/kk u igqapk;sA ok;q esa /ofu dh pky 320 m/s. [ 3 min.] [M.Bank(07-08)_HCV_Ch.16__Ex._2] Ans. 0.5 s Sol.
t=
80 2 = 0.5 sec, 320
Prayer meeting
Building 80 m
3.
A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air. ,d O;fDr] ,d cgqr cM+h nhokj ls 50.0 eh- nwj [kM+k gqvk gS] ,oa fu;fer varjky ls rkyh ctk jgk gSA izkjEHk esa
varjky vf/kd gSA og /khjs&/khjs varjky de djrk gSA rFkk bldks bruk fu;r dj nsrk gS fd ,d rkyh dh vuqxawt vxyh rkyh ds lkFk lEikfrr gks tkrh gSA ;fn mldks izR;sd 3 lsd.M esa 10 ckj rkyh ctkuh iM+ jgh gks] ok;q esa /ofu dk osx Kkr dhft;sA Ans. 333 m/s 4.
A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m/s. ,d O;fDr 20 gVZ~t ls 20 fdyks gVZ~t vko`fÙk ijkx dh /ofu rjaxs lqu ldrk gSA O;fDr dks lqukbZ ns ldus okyh vf/ kdre rFkk U;wure /ofu rjaxnS/;Z Kkr dhft;sA /ofu dh pky 360 eh@ls gSA Ans. 18 mm, 18 m
5.
Find the minimum and maximum wavelengths of sound in water that is in the audible range (20 - 20000 Hz) for an average human ear. Speed of sound in water = 1450 m/s. ,d vkSlr euq"; ds dku dks lqukbZ nsus okyh JO; ijkl (20 - 20000 gV~Zt) dh /ofu rjaxks dh ikuh esa vf/kdre rFkk U;wure /ofu rjaxnS/;Z Kkr dhft;sA ?ofu dh pky = 1450 m/s. Ans. 7.25 cm, 72.5 m
6.
Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker. (a)Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm. (b) Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above. Take the speed of sound to be 340 m/s.
;fn /ofu rjaxks dh rjaxnS/;Z ykmM Lihdj ds O;kl ls cgqr vf/kd gks rks ykmM Lihdj ls /ofu rjaxsa] leLr fn'kkvks manishkumarphysics.in
Page # 13
Chapter # 16
Sound Waves
esa ,d leku :i ls QSyrh gSA(a) ml vko`fÙk dh x.kuk dhft;s ftlds fy;s /ofu dh rjaxnS/;Z Lihdj ds O;kl dh nl xquh gS] Lihdj dk O;kl 20 lseh gSA (b) ;fn rjaxnS/;Z] ykmM Lihdj ds O;kl ls cgqr de gks rks /ofu fuf'pr :i ls dsoy vkxs dh fn'kk esa gh LFkkukarfjr gksrh gSA ml vko`fÙk dh x.kuk dhft;s ftldh rjaxnS/;Z ykmM Lihdkj ds O;kl dh nloka Hkkx gksA /ofu dh pky 340 eh-@ls eku yhft;sA Ans. (a) 170 Hz (b) 17 kHz 7.
Ultrasonic waves of frequency 4.5 MHz are used to detect tumour in soft tissues. The speed of sound in tissue is 1.5 km/s and that in air is 340 m/s. Find the wavelength of this ultrasonic wave in air and in tissue. dksey Årdks esa V~;wej dh tkap ds fy;s 4.5 esxk gV~Zt vko`fÙk dh ijkJO; rjaxs iz;qDr dh tkrh gSA Ård esa /ofu dh pky 1.5 fdeh@ls rFkk ok;q esa 340 eh-@ls gSA ok;q rFkk Ård esa ijkJO; rjax dh rjaxnS/;Z Kkr dhft;sA Ans. 7.6 × 10–5 m, 3.3 × 10–4 m
8.
The equation of a travelling sound wave is y = 6.0 sin (600 t – 1.8 x) where y is measured in 10–5 m, t in second and x in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed. ,d izxkeh /ofu rjax dh lehdj.k y = 6.0 sin (600 t – 1.8 x) gS] tgk¡ y, 10–5 eh- es t lsd.M esa rFkk x ehVj esa ekis tkrs gSA (a) d.kkas ds foLFkkiu vk;ke rFkk rjax dh rjaxnS/;Z dk vuqikr Kkr dhft;sA (b) d.kksa ds osx vk;ke rFkk
rjax pky dk vuqikr Kkr dhft;sA
Ans. (a) 1.7 × 10–5 m, 3.3 × 10–4 m 9.
A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m/s. (a) By how much is the phase changed at a given point in 2.5 ms? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation ? 100 gVZ~t vko`fÙk okyh /ofu rjax ok;q esa xeu dj jgh gSA ok;q esa /ofu dh pky 350 eh-@ls (a) fn;s x;s fdlh fcUnq ij 2.5 ms (feyh lsd.M) (b) xeu dh fn'kk ds vuqfn'k 10.0 lseh ij fLFkr nks fcUnqvksa ij fdlh {k.k fdruk dykUrj
gks xk\
Ans. (a) /2
(b) 2/35
10.
Two points sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources? /ofu ds nks fcUnq L=kksr 10 lseh nwjh ij j[ks gq, gSA os leku dyk esa dEiUu djrs gq, 5.0 lseh rjaxnS/;Z okyh rjaxs mRiUu djrs gSA ,d L=kksr ls 20 lseh nwj fLFkr fcUnq ij igqpus okyh nks rjaxks esa fdruk dykarj gksxk& (a) L=kksr dks feykus okyh js[kk ij ,oa (b) nksuksa L=kksrksa dks feykus okyh js[kk ds yEc v/kZd ij Ans. (a) zero (b) zero
11.
Calculate the speed of sound in oxygen from the following data. The mass of 22.4 litre of oxygen at STP (T = 273 K and p= 1.0 × 105 N/m2) is 32 g, the molar heat capacity of oxygen at constant volume is Cv = 2.5 R and that at constant pressure is Cp = 3.5 R. fuEufyf[kr vkadM+ksa ds vk/kkj ij vkDlhtu esa /ofu dh pky dh x.kuk dhft;sA STP (T = 273 K rFkk p= 1.0 × 105 N/m2) ij 22.4 yhVj vkDlhtu dk nzO;eku 32 xzke gksrk gS] fu;r vk;ru ij vkDlhtu dh eksy j fof'k"V m"ek Cv = 2.5 R ,oa fu;r nkc ij Cp = 3.5 R Ans. 310 m/s
12.
The speed of sound as measured by a student in the laboratory on a winter day is 340 m/s when the room temperature is 17°C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32°C? [HCV_Ch.16__Ex._12] lnhZ ds fnuksa es tc dejs dk rki 17ºC gS] fdlh fo|kFkhZ }kjk /ofu dh pky 340 m/s ekih xbZA dksbZ fo|kFkhZ blh iz;ksx dks nksgjkrk gS rc dejs dk rki 32°C gksrk gS] og fdruh pky ekisxk \ Ans. 349 m/s
13.
At what temperature will the speed of sound be double of its value at 0°C? fdl rki ij /ofu dh pky] blds 0°C ij eku dh nwxuh gks tk;sxh \ [HCV_Ch.16__Ex._13] Ans. 819°C
14.
The absolute temperature of air in a region linearly increases from T1 to T2 in a space of width d. Find the time taken by a sound wave to go through the region in terms of T1, T2 and the speed v of sound at 273 K. Evaluate this time for T1 = 280 K, T2 = 310 K, d = 33 m and v = 330 m/s. manishkumarphysics.in Page # 14
Chapter # 16
Sound Waves
vkdk'k dh d pkSM+kbZ okys fdlh {ks=k esa ok;q dk ijerki T1 ls T2 rd jSf[kd :i ls c<+rk gSA 273 K rki ij bl {ks=k ls xqtjus esa /ofu rjax }kjk fy;k x;k le; T1, T2 d ,oa /ofu dh pky v ds inks esa Kkr dhft;sA T1 = 280 K, T2 = 310 K, d = 33 m ,oa v = 330 m/s ds fy;s le; dk eku izkIr dhft;sA Ans.
2d .
273 T1 T2
, 96 ms
15.
Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of 2.0 × 105 N/m2 from the following data. Density of kerosene = 800 kg/m3 and speed of sound in kerosene is 1330 m/s. fuEu vkadM+ks ds vk/kkj ij 1.0 yhVj dSjksflu ij vfrfjDr nkc 2.0 × 105 N/m2 vkjksfir djus ls vk;ru esa ifjorZu Kkr dhft;sA dSjksflu dk ?kuRo = 800 kg/m3 rFkk dSjksflu esa /ofu dh pky 1330 m/s. Ans. 0.14 cm3
16.
Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm travelling in air. The pressure at a point varies between (1.0 × 105 ± 14) Pa and the particles of the air vibrate in simple harmonic motion of amplitude 5.5 × 10–6 m. ok;q esa xeu dj jgh 35 lseh rjaxnS/;Z okyh /ofu rjax ds fy;s fn;s x;s fuEu vkadM+ks ds vk/kkj ij ok;q ds vk;ru izR;kLFkrk xq.kkad dh x.kuk dhft;sA fdlh fcUnq ij nkc (1.0 × 105 ± 14) ikLdy ds chp ifjofrZr gksrk gSA ,oa ok;q ds d.k ljy vkorZ xfr es 5.5 × 10–6 m vk;ke ds dEiUu djrs gSA Ans. 1.4 × 105 N/m2
17.
A source of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/s and the density of air is 1.2 kg/m3. (a) What is the intensity at a distance of 6.0 m from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point? 2.0 fdyks gV~Zt 20 okWV ij izpfyr ,d /ofu L=kksr leLr fn'kkvksa esa leku :i ls /ofu mRlftZr djrk gSA ok;q esa /ofu pky 340 eh@ ls rFkk ok;q dk ?kuRo 1.2fdxzk@eh-3 (a) L=kksr ls 6.0 eh nwj rhozrk fdruh gksxh\ (b) bl fcUnq ij nkc vk;ke fdruk gksxk\ (c) bl fcUnq ij foLFkkiu vk;ke fdruk gksxk\ Ans. (a) 44 mW/m2 (b) 6.0 Pa (c) 1.2 × 10–6 m
18.
The intensity of sound from a point source is 1.0 × 10–8 W/m2 at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source? ,d fcUnq L=kksr ls 50 eh nwj fLFkr fdlh fcUnq ij /ofu dh rhozrk 1.0 × 10–8 okWV@eh2 gSA L=kksr ls 25 eh nwj rhozrk
fdruh gksxh\
Ans. 4.0 × 10–10 W/m2 19.
The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source? ,d fcUnq L=kksr ls 5.0 eh nwj /ofu dk Lrj 40 Mslhcy (dB) gSA L=kksr ls 50 eh- nwj Lrj fdruk gksxk\ Ans. 20 dB
20.
If the intensity of sound is doubled, by how many decibels does the sound level increase?
;fn /ofu dh rhozrk nqxuh dj nh tk;s /ofu dk Lrj fdrus Mslhcy c<+ tk;sxk\ Ans. 3 dB 21.
Sound with intensity larger than 120 dB appears painful to a person. A small speaker deliveries 2.0 W of audio output. How close can the person get to the speaker without hurting his ears? 120 dB ls vf/kd rhozrk okyh /ofu fdlh O;fDr ds fy;s vlguh; gksrh gSA ,d NksVk Lihdj 2.0 okWV /ofu mRlftZr
djrk gSaA dksbZ O;fDr fcuk dkuksa dks uqdlku igqapk;s] Lihdj ds fdruk lehi vk ldrk gSA Ans. 40 cm 22.
If the sound level in a room is increased from 50 dB to 60dB, by what factor is the pressure amplitude increased? ;fn fdlh dejs esa /ofu dk Lrj 50 dB ls 60dB c<+k fn;k tk;s] nkc vk;ke fdl xq.kd ls c<+ tk;sxk\ Ans. 10 manishkumarphysics.in
Page # 15
Chapter # 16 Sound Waves 23. The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student output same sound energy per second, what will be the noise level if the number of students is increased to 100 ? fdlh d{kk esa v/;kid dh vuqifLFkfr esa 'kksj dk Lrj 50 dB gS rFkk 50 fo|kFkhZ mifLFkr gSA eku fyft;s fd izR;sd fo|kkFkhZ vkSlru leku ÅtkZ izfrlsd.M fuxZfer djrk gS] ;fn fo|kfFkZ;ksa dh la[;k c<+kdj 100 dj nh tk;s rks 'kksj dk Lrj fdruk gksxk ? Ans. 53 dB 24.
In a Quincke’s experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m/s. fDoad ds iz;ksx esa f[kldus okyh uyh dks 2.50 lseh f[kldkus ij lalwfpr /ofu mPpre ls u;wure gks tkrh gSA ;fn ok;q esa /ofu dh pky 340 eh-@ls- gS] /ofu dh vko`fÙk Kkr dhft;sA Ans. 3.4 kHz
25.
In a Quincke’s experiment, the sound intensity has a minimum value at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm, the intensity increases to a maximum of 9. Take the speed of sound in air to be 330 m/s. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity. fDoad iz;ksx esa] fdlh fo'ks"k fLFkfr esa /ofu dh rhozrk dk U;wure eku gSA tSls gh f[kldus okyh uyh dks 16.5 feeh ckgj [khpk tkrk gS] rhozrk c<+dj vf/kdre 9 gks tkrh gSA ok;q esa /ofu dh pky 330 eh@ls ekudj (a) /ofu L=kksr dh vko`fr Kkr dhft;sA (b) ;g ekurs gq, fd vf/kdre rFkk U;wure rhozrkvksa ds chp vk;ke esa dksbZ fo'ks"k ifjorZu
ugh gksrk gSA lalwpd rd vkus okyh nksuksa rjaxks ds vk;keksa dk vuqikr Kkr dhft;sA Ans. (a) 5.0 kHz
(b) 2
26.
Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a distractive interference at the place of the listener? Speed of sound in air = 320 m/s. nks /ofu Lihdj dqN nwjh ij j[ks gq, gS rFkk ,d gh izo/kZd ls ctk;s tkrs gSA ,d O;fDr fdlh ,d Lihdj ls 6.0 eh- rFkk nwljs ls 6.4 eh- nwj cSBk gqvk gSA ;fn /ofu ladsr fujUrj 500 gVZ~t ls 5000 gV~Zt ds chp ifjofrZr gksrk gSA Jksrk dh fLFkfr ij fdu vko`fÙk;ksa ds fy;s fouk'kh O;fDrdj.k gksxk\ ok;q esa /ofu dh pky = 320 eh-@lsAns. 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz
27.
A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. Velocity of sound in air is 336 m/s. ,d /ofu L=kksr rFkk llap w d D ijLij dqN nwjh ij j[ks gq, gSA llap q d ds lehi cM+k dkMZ cksMZ D, js[kk SD ds yEcor~ fp=kuqlkj j[kk gqvk gSA ;g /khjs&/khjs nwj fLFkkfir fd;k tkrk gS rFkk ;g izf{kr gksrk gSA tc cksMZ dks 20 lseh foLFkkfir
fd;k tkrk gSA rhozrk dk eku vf/kdre ls U;wure rd ifjofrZr gks tkrk gSA mRlftZr /ofu dh vko`fÙk dhft;sA ok;q esa /ofu dh pky 336 eh-@ls-
Ans. 420 Hz 28.
A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance H from the source and the detector as shown in figure. The source emits a wave of wavelength = d/2 which is received by the detector after reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direction wave? ,d L=kksr S rFkk lalwpd D, d nwjh ij j[ks gq, gSA fp=kkuqlkj ,d cM+h dkMZcksMZ L=kksr rFkk lalwpd ls H nwjh ij j[kk gqvk gSA L=kksr ls mRlftZr rjax dh rjax nS/;Z = d/2 gS] tks dkMZcksMZ ls ijkorZu ds i'okr~ lalwpd }kjk xz.k dh tkrh
gSA ;g L=kksr ls lh/kh izkIr gksus okyh rjaxksa ds leku dyk esa gh gSA dkMZ cksM+Z dks U;wure fdl nwjh ls f[kldk;k tk;s fd ijkofrZr rjax] lh/kh rjax ls foijhr dyk esa gks tk;s \ Ans. 0.13 d manishkumarphysics.in
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Chapter # 16
Sound Waves
29.
Two stereo speakers are separated are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure. Find the frequencies in the audible range (20-20000 Hz) for which the listener will hear a minimum sound intensity. Speed of sound in air = 320 m/s. nks LVhfj;ksa Lihdjksa ds e/; 2.40 eh- nwjh gSA ,d O;fDr fp=k esa n'kkZ;s vuqlkj ,d lzksr ls lh/kk 3.20 eh- nwj [kM+k gqvk gSA JO; ijkl esa (20 - 20000 gV~Zt) og vko`fÙk;k¡ Kkr dhft;s] ftuds fy;s Jksrk U;wure /ofu rhozrk lqusxkA ok;q esa /ofu dh pky = 320 eh@ls- A
Ans.
200 (2n + 1) Hz where n = 0, 1, 2, ....... 49
30.
Two sources of sound, S1 and S2, emitting waves of equal wavelength 20.0 cm, are placed with a separation of 20.0 cm between them. A detector can be moved on a line parallel to S1S2 and at a distance of 20.0 cm from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound. nks /ofu lzksr S1 rFkk S2, leku rjaxnS/;Z 20.0 lseh dh /ofu mRlftZr dj jgs gSa] 20.0 lseh nwjh ij j[ks gq, gSaA S1S2 js[kk ds lekUrj ,d lalwpd xfr dj ldrk gS] ;g blls 20.0 lseh nwj gSA izkjEHk esa lalwpd nksuksa lzksrksa ls leku
nwjh ij gA eku yhft;s fd lzksrksa ls mRlftZr rjaxsa leku dyk esa gSA Kkr dhft;s fd U;wure /ofu lalwfpr djus ds fy;s lalwpd dks U;wure fdruh nwj foLFkkfir djuk iM+sxkA Ans. 12.6 cm 31.
Two speakers, S1 and S2, driven by the same amplifier, are placed at y = 1.0 m and y = – 1.0 m (figure). The speakers vibrate in phase at 600 Hz. A man stands at a point on the X-axis at a very large distance from the origin and starts moving parallel to the Y-axis. The speed of sound in air is 330 m/s. (a) At what angle will the intensity of sound drop to a minimum for the first time? (b) At what angle will he hear a maximum of sound intensity for the first time? (c) If the continues to walk along the line, how many more maxima can he hear? ,d gh izo/kZd }kjk pkfyr nks Lihdj S1 o S2 , y = 1.0 eh ,oa y = – 1.0 eh (fp=k) Lihdj leku dyk esa 600 gVZ~t ls dEiu dj jgs gSaA ,d O;fDr X-v{k ij ewy fcUnq ls cgqr nwj [kM+k gqvk gS rFkk Y-v{k ds lekukUrj pyuk 'kq: djrk gSA ok;q esa /ofu dh pky 330 eh@ls gSA (a) izFke ckj fdl dks.k ij /ofu dh rhozrk 'kwU; gksxh\ (b) og izFke ckj fdl dks.k ij vf/kdre /ofu rhozrk lqusxk\ (c) ;fn og bl js[kk ds vuqfn'k pyrk jgrk gS] og fdrus mfPp"B
vkSj lqusxk\
Ans. (a) 7.9°
(b) 16°
(c) two
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Chapter # 16 Sound Waves 32. Three sources of sound S 1, S 2 and S 3 of equal intensity are placed in a straight line with S1S2 = S2S3 (figure). At a point P, far away from the sources, the wave coming from S2 is 120° ahead in phase of that from S1. Also, the wave coming from S3 is 120° ahead of that from S2. What would be the resultant intensity of sound at P? leku rhozrk okys rhu /ofu lzksr S1, S2 rFkk S3 ,d ljy js[kk ds vuqfn'k j[ks gq, gSa ,oa S1S2 = S2S3 gSa (fp=k). At a point P, far away from the sources, the wave coming from S2 is 120° ahead in phase of that from S1. Also, the wave coming from S3 is 120° ahead of that from S2. What would be the resultant intensity of sound at P?
k
Ans. zero 33.
Two coherent narrow slits emitting sound of wavelength l in the same phase are placed parallel to each other at a small separation of 2. The sound is detected by moving a detector on the screen S at a distance D (>> l) from the slit S1 as shown in figure. Find the distance x such that the intensity at P is equal to the intensity at O. k Two coherent narrow slits emitting sound of wavelength l in the same phase are placed parallel to each other at a small separation of 2. The sound is detected by moving a detector on the screen S at a distance D (>> l) from the slit S1 as shown in figure. Find the distance x such that the intensity at P is equal to the intensity at O. k
Ans. 34.
3D
Figure shows two coherent sources S1 and S2 which emit sound of wavelength l in phase. The separation between the sources is 3l. A circular wire of large radius is placed in such a way that S1 S2 lies in its plane and the middle point of S1 S2 is at the centre of the wire. Find the angular positions q on the wire for which constructive interference takes place. k Figure shows two coherent sources S1 and S2 which emit sound of wavelength l in phase. The separation between the sources is 3l. A circular wire of large radius is placed in such a way that S1 S2 lies in its plane and the middle point of S1 S2 is at the centre of the wire. Find the angular positions q on the wire for which constructive interference takes place. k
Ans. 0°, 48.2°, 70.5°, 90° and similar points in other quadrants 35.
Two sources of sound S1 and S2 vibrate at same frequency and are in phase (figure). The intensity of sound detected at a point P as shown in the figure is 0. (a) if equals 45°, what will be the intensity of sound detected at this point if one of the sources is switched off? (b) What will be the answer of the previous part if = 60°? k Two sources of sound S1 and S2 vibrate at same frequency and are in phase (figure). The intensity of sound detected at a point P as shown in the figure is 0. (a) if equals 45°, what will be the intensity of sound detected at this point if one of the sources is switched off? (b) What will be the answer of the previous part if = 60°? k
Ans. (a) I0/4 (b) I0/4 manishkumarphysics.in
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Chapter # 16 Sound Waves 36. Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is 340 m/s. 20 lseh yEckbZ okys vkxZu ikbi ds fy;s ewy Loj] izFke lUukn ,oa f}rh; lUukn dh vko`fÙk;k¡ Kkr dhft;sA ok;q esa /ofu dh pky 340 eh@ls Ans. 850 Hz, 1700 Hz and 2550 Hz 37.
A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air = 340 m/s. ,d cUn vkxZu ikbi U;wure 500 gVZ~t vko`fÙk ls dEiUu dj ldrk gSA ikbi dh yEckbZ Kkr dhft;sA ok;q esa /ofu dk osx = 340 eh@ls. Ans. 17 cm
38.
In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328 m/s, what is the frequency of the source? dEiUu'khy ok;q LrEHk esa mRiUu vizxkeh rjax izfr:i esa] 4.0 lseh nwjh ij fuLian curs gSaA ;fn ok;q esa /ofu dh pky 328 eh@ls gks] lzksr dh vko`fÙk fdruh gS\ Ans. 4.1 kHz
39.
The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m/s, find the frequency of vibration of the air column. dEiu'khy ok;q LrEHk esa fdlh foLian rFkk mlls vxys izLian ds chp dh nwjh 25 lseh gSA ;fn ok;q esa /ofu dh pky 340 eh@ls gS] ok;q LrEHk ds dEiUu dh vko`fÙk Kkr dhft;sA Ans. 340 Hz
40.
A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. Speed of sound in air is 340 m/s. /kkrq dh ,d csyukdkj uyh dh yEckbZ 50 lseh gS rFkk ;g nksuksa fljksa ij [kqyh gqbZ gSA 1000 Hz rFkk 2000 Hz ds chp dh vko`fÙk;k¡ Kkr dhft;s ftlds fy;s uyh esa ok;q LrEHk vuqukfnr gks ldrk gSA ok;q esa /ofu dh pky 340 eh@ls
gSA
Ans. 1020 Hz, 1360 Hz and 1700 Hz 41.
In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form? vuqukn LrEHk iz;ksx esa] 400 gVZ~t vko`fÙk okys Lofj=k f}Hkqt iz;qDr fd;k tkrk gSA LrEHk dh yEckbZ 20.0 lseh gksus ij izFke vuqukn izsf{kr fd;k tkrk gS rFkk f}rh; vuqukn ok;q LrEHk dh 62.0 lseh yEckbZ ds fy;s izsf{kr gksrk gSA (a) ok;q esa /ofu dh pky Kkr dhft;s (b) [kqy s fljs ls fdruh nwj Åij nkc fuLian cusxk\ Ans. (a) 336 m/s (b) 1 cm
42.
The first overtone frequency of a closed organ pipe P1 is equal to the fundamental frequency of an open organ pipe P2. If the length of the pipe P1 is 30 cm, what will be the length of P2? ,d cUn vkxZuikbi P1 dh izFke lUukn vko`fÙk] ,d [kqys ikbi P2 dh ewy vko`fÙk ds cjkcj gSA ;fn ikbi P1 dh yEckbZ 30 lseh gS] P2 dh yEckbZ fdruh gksxh ? Ans. 20 cm
43.
A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz-20,000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km/s. rkacs dh 1.0 eh yEch NM+ e/; esa dlh gqbZ gSA 20 gVZ~t ls 20,000 gVZ~t ds e/; rjaxksa dh vko`fÙk;k¡ Kkr dhft;s] ftuds fy;s NM+ esa vuqnS/;Z vizxkeh rjaxsa cu ldsA rkacs esa /ofu dh pky 3.8 fdeh@ls Ans. 1.9 n kHz where n = 1, 2, 3, ........, 10
44.
Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20-20,000 Hz). Speed of sound in air = 340 m/s.
nksuksa fljksa ij [kqys gq, fdlh vkxZu ikbi dh vf/kdre yEckbZ Kkr dhft;s] ftldh ewy vko`fÙk lkekU; Jo.k ijkl (20 ls 20,000 gVZ~t) esa gksA ok;q esa /ofu dh pky = 340 eh@ls Ans. 8.5 m
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Chapter # 16 Sound Waves 45. An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) what is the highest harmonic of such a tube that is in the audible range is in the audible range ? speed of sound in air is 340 ms/ and the audible range is 20-20,000 Hz. ,d [kqys vkxZu ikbi dh yEckbZ 5 lseh gSA (a) bl ikbi ds dEiUu dh ewy vko`fÙk Kkr dhft;sA (b) bl ikbi dk dkSulk mPpre lUukfn JO; ijkl esa gS\ ok;q esa /ofu dh pky 340 eh@ls rFkk Jo.k ijkl 20 - 20,000 gVZ~t gSA Ans. (a) 3.4 kHz (b) 5 46.
An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz2kHz. Find the frequencies at which the column will resonate. Speed of sound in air = 320 m/s.
bysDVªkWfud pkfyr ,d ykmM Lihdj.k] vuqukn LrEHk midj.k ds lehi j[kk gqvk gSA uyh esa ok;q LrEHk dh yEckbZ 80 lseh gSA ykmM&Lihdj dh vko`fÙk 20 gVZ~t&2 fdyks gVZ~t ijkl esa ifjofrZr dh tk ldrh gSA og vuqukfnr gksxkA ok;q esa /ofu dh pky = 320 eh@ls Ans. 100 (2n + 1) Hz where n = 0, 1, 2, 3, ...., 9 47.
Two successive resonance frequencies in an open organ pipe are 1944 and 2592 Hz. Find the length of the tube. The speed of sound in air is 324 m/s. fdlh [kqys vkxZu esa nks Øekxr vuqukfn vko`fÙk;k¡ 1944 gVZ~t ,oa 2592 gVZ~t gSA uyh dh yEckbZ Kkr dhft;sA ok;q esa /ofu dh pky 324 eh@ls gSA Ans. 25 cm
48.
A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out of the tube through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube. de vuqizLFk dkV okyh ,d uyh esa fiLVu yxk gqvk gS ,oa bldk nwljk fljk gSA uyh 512 gVZ~t vko`fÙk okys Lofj=k
f}Hkqt ds lkFk vuqukfnr gksrh gSA fiLVu dks /khjs&/khjs ckgj [khapk tkrk gS rFkk ;g izsf{kr fd;k tkrk gS fd tc fiLVu 32.0 lseh ckgj [khapk x;k gS] f}rh; vuqukn mRiUu gksrk gSA uyh dh ok;q esa /ofu dh pky dh x.kuk dhft;sA Ans. 328 m/s 49.
A U-tube having unequal arm-lengths has water in it. A tuning fork of frequency 440 Hz can set up the air in the shorter arm in its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibration. Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air = 330 m/s. ,d vleku yEckbZ dh Hkqtkvksa okyh U-uyh esa ikuh Hkjk gqvk gSA 440 gVZ~t vko`fÙk okys Lofj=k f}&Hkqt NksVh Hkqtk
dh ok;q esa ewy Loj ds dEiUu mRiUu djrk gS rFkk ;gh Lofj=k f}Hkqt yEch Hkqtk dh ok;q esa izFke lUukn ds dEiUu mRiUu djrk gSA ok;q LrEHkksa dh yEckbZ;k¡ Kkr dhft;sA vUR; izHkko ux.; eku yhft;s ,oa ok;q esa /ofu dh pky = 330 eh@ls 50.
Ans. 18.8 cm, 56.3 cm Consider the situation shown in figure. The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 m/s, find the tension in the wire. fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA rkj dk nzO;eku 4.00 xzke gS ,oa ;g f}rh; lUukfn esa dEiUu dj jgk gS rFkk ;g uyh ds ok;q LrEHk dks ewy fo/kk esa dfEir djrk gSA ok;q esa /ofu dh pky 340 eh@ls eku yhft;s] rkj dk
ruko Kkr dhft;sA
Ans. 11.6 N
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Chapter # 16 Sound Waves 51. A 30.0 cm long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0 cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = 340 m/s. ,d 30.0 lseh yEck rkj nksuksa fljksa ij dldj rkuk x;k gS] bldk nzO;eku 10.0 xzke gS ,oa ;g ewy fo/kk esa dfEir djk;k x;k gSA ,d 50.0 lseh yEcs can vkxZu ikbi dk [kqyk fljk rkj ds lehi j[kk gqvk gS] rkj dks dfEir djkus ij ikbi esa ewy fo/kk ds dEiUu gksrs gSaA rkj esa ruko Kkr dhft;sA ok;q esa /ofu dh pky = 340 eh@ls Ans. 347 N 52.
Show that if the room temperature changes by a small amount from T to T + T, the fundamental frequency of an organ pipe changes from v to v + v, where O;Dr dhft;s fd ;fn dejs dk rki esa T ls vR;Yi ifjofrZr gksdj T + T, gks tkrk gS] rks fdlh vkxZu ikbi dh ewy vko`fÙk ifjofrZr gksdj v ls v + v, gks tkrh gS] tgk¡
v 1 T = . v 2 T 53.
The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20°C. What will be its fundamental frequency when the temperature changes to 22°C? fdlh can ikbi esa tc ok;q dk rki 20°C gS] bldh ewy vko`fÙk 293 gVZ~t gSA bldk rkj ifjofrZr gksdj 22°C gksus ij bldh ewy vko`fÙk fdruh gksxh ? Ans. 294 Hz
54.
A Kundt’s tube apparatus has a copper rod of length 1.0 m clamped at 25 cm from one of the ends. The tube contains are in which the speed of sound is 340 m/s. The powder collects in heaps separated by a distance of 5.0 cm. Find the speed of sound waves in copper. ,d dq.M uyh iz;ksx midj.k 1.0 eh yEch rkacs dh NM+ ,d fljs ls 25 lseh nwjh ij dlh gqbZ gSA uyh esa ok;q Hkjh gqbZ gS] ftlesa /ofu dh pky 340 eh@ls gSA ,df=kr ikmMj dh
dh pky Kkr dhft;sA Ans. 3400 m/s 55.
A Kundt’s tube apparatus has a steel rod of length 1.0 m clamped at the centre. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in air. dq.M uyh iz;ksx midj.k esa 1.0 eh yEch LVhy dh NM+ blds dsUnz ij dlh gqbZ gSA NM+ dks bldh ewy fo/kk esa 2600 gVZ~t vko`fÙk ls dfEir djok;k tkrk gSA uyh esa fc[ksjk x;k ykbdksiksfM;e ikmMj 6.5 lseh dh nwjh ij fLFkr
NksVh&NksVh
A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476 hz. or 480 Hz. What is the frequency of the tuning fork? ,d ifjorZu'khy vko`fÙk okyk /ofu lzksr tc bldh vko`fÙk 476 gVZ~t ;k 480 gVZ~t gksrh gS rks ,d Lofj=k gVZ~t ds lkFk 2 foLian mRiUu djrk gSA Lofj=k f}&Hkqt dh vko`fÙk fdruh gS\ Ans.478 Hz
57.
A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork? ,d Lofj=k f}Hkqt] 256 gVZ~t vko`fÙk okys ,d vU; Lofj=k f}Hkqt ds lkFk 4 foLian izfr lsd.M mRiUu djrk gSA izFke Lofj=k f}Hkqt ij FkksM+k lk ekse yxkus ij] foLian vko`fÙk c<+dj 6 izfr lsd.M gks tkrh gSA Lofj=k f}Hkqt dh ewy vko`fÙk
fdruh Fkh\
Ans.252 Hz 58.
Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of 32 cm and the other of 32.2 cm. The speed of sound in air is 350 m/s. tc nks /ofu lzksrksa dks ok;q esa ,d lkFk dfEir djk;k tkrk gS] ,d lzksr 32 lseh rFkk nwljk 32.2 lseh rjaxnS/;Z dh /ofu mRlftZr djrk gS] foLian vko`fÙk dh x.kuk dhft;sA ok;q esa /ofu dh pky 350 eh@ls A Ans. 7 per second
59.
A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m/s. ,d vKkr vko`fÙk dk Lofj=k f}Hkqt ,d vU; Lofj=k ds lkFk 5 foLian izfr lsd.M mRiUu djrk gS] tks 40 lseh yEckbZ manishkumarphysics.in
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Chapter # 16
Sound Waves
okys can vkxZu ikbi esa bldh ewy fo/kk esa dEiUu mRiUu djrk gSA tc izFke Lofj=k f}Hkqt ij FkksM+k lk ekse yxk;k tkrk gS] foLian dh vko`fÙk de gks tkrh gSA okLrfod vko`fÙk Kkr dhft;sA ok;q esa /ofu dh pky 320 eh@ls gSA Ans. 205 Hz 60.
A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B. ,d fi;kuksa rkj A , 600 gVZ~t ewy vko`fÙk esa dEiUu djrk gSA ,d nwljk ,slk gh rkj B , 6 foLian izfr lsd.M mRiUu djrk gS] tc A dk ruko FkksM+k lk c<+k fn;k tk;sA A o B esa rukoksa dk vuqikr Kkr dhft;sA Ans. 1.02
61.
A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the height of slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork? 256 gVZ~t vko`fÙk okyk ,d Lofj=k f}Hkqt ewy fo/kk esa dEiUu xdj jgs 25 lseh yEcs rkj ds lkFk 4 foLian izfr lsd.M
mRiUu dj jgk gSA tc yEckbZ FkksM+h lh de dj nh tkrh gS] foLian vko`fÙk de gks tkrh gSA rkj dh yEckbZ fdruh de dh tk ldrh gS fd ;g Lofj=k f}Hkqt ds lkFk dksbZ foLian mRiUu ugha djsaA Ans. 0.39 cm 62.
A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the apparent frequency heard by a scooter-driver approaching the policeman at a speed of 36.0 km/h? Speed of sound in air = 340 m/s. lM+d ij [kM+k gqvk VªsfQd iqfyl dk flikgh lhVh ctkrk gSA ftldh eq[; vko`fÙk 2.00 fdyks gV~Zt gSA iqfyl flikgh dh vksj 36.0 fdeh@?kaVk pky ls vk jgk LdwVj pkyd fdruh vkHkklh vko`fÙk lqusxk\ ok;q esa /ofu dh pky = 340 eh@lsA Ans. 2.06 kHz
63.
The horn of a car emits sound with a dominant frequency of 2400 Hz. What will be the apparent dominant frequency heard by a person standing on the road is front of the car if going at 18.0 km/h? Speed of sound in air = 340 m/s. ,d dkj ds gkWuZ dh eq[; /ofu dh vko`fÙk 2400 gVZ~t gSA ;fn dkj 18.0 fdeh@?kaVk pky ls tk jgh gS rks dkj ds lkeus lM+d ij [kM+k gqvk O;fDr eq[; /ofu dh vkHkklh vko`fÙk fdruh lqusxk\ ok;q esa /ofu dh pky = 340 eh@ls Ans. 2436 Hz
64.
A person riding a car moving at 72 km/h sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (A) in front of the car (b) behind the car? Speed of sound in air = 340 m/s. 72 fdeh@?kaVk pky ls xfr'khy dkj esa lokj ,d O;fDr 1250 gVZ~t vko`fÙk dh lhVh ctkrk gSA ,d vU; O;fDr fdruh vko`fÙk lqusxk] ;fn og [kM+k gSA (A) dkj ds lkeus (b) dkj ds ihNs\ ok;q esa /ofu dh pky = 340 eh@ls Ans. (a) 1328 Hz (b) 1181 Hz
65.
A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m/s. IysVQkeZ dh vksj 54 fdeh@?kaVk pky ls vk jgh ,d Vªsu lhVh ctkrh gSA IysVQkeZ ij [kM+k gqvk ,d izs{kd bldh vko`fÙk 1620 gVZ~t lqurk gSA lhVh ctkrh gqbZ Vªsu fcuk /kheh gq, IysVQkeZ ls xqtj tkrh gSA Vªsu ds IysVQkeZ ikj djus ds i'pkr~ izs{kd fdruh vko`fÙk lqusxk\ ok;q esa /ofu dh pky = 332 eh@lsA Ans. 1480 Hz
66.
A bat emitting an ultrasonic wave of frequency 4.5 × 104 Hz flies at a speed of 6 m/s between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m/s. nks lekukarj nhokjksa ds chp 6 eh@ls pky ls mM+ jgk ,d pexknM+ 4.5 × 104 gVZ~t vko`fÙk dh ijkJO; rjax mRiUu
dj jgk gSA pexknM+ }kjk lquh x;h nks vko`fÙk;k¡ rFkk nksuksa ds e/; foLian vko`fÙk Kkr dhft;sA ok;q esa /ofu dh pky 330 eh@ls gSA Ans. 4.67 × 104 Hz, 4.34 × 104 Hz, 3270 Hz 67.
A bullet passes past a person at a speed of 220 m/s. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air = 330 m/s. ,d O;fDr ds ikl ds canwd dh xksyh 220 eh@ls pky ls xqtj tkrh gSA tSls gh xksyh O;fDr ds ikl ls xqtjrh gS]
xksyh ds dkj.k mRiUu lhVh dh /ofu dh vko`fÙk esa fdl ?kVd ls ifjorZu lquk;h nsxk] Kkr dhft;sA ok;q esa /ofu manishkumarphysics.in
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dh pky = 330 eh@lsA Ans.0.8 68.
Two electric trains run at the same speed of 72 km/h along the same track and in the same direction with a separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whisting. If both the whistles were at 500 Hz and the speed of sound in air is 340 m/s, find the frequencies heard by the person. ,d gh iFk ij nks fo|qr Vªsusa] ,d gh fn'kk esa 72 fdeh ?kaVk dh leku pky ls xfr'khy gS] buds e/; dh nwjh 2.4
fdeh gSA nksuksa Vªsus ,d lkFk NksVh lh lhVh ctkrh gSA ftl {k.k ;g lhVh ctkrh gS] ,d O;fDr jsy iFk ls yEcor~ 500 eh nwj rFkk nksuksa Vªsuksa ls leku nwjh ij [kM+k gqvk gSA ;fn nksu ksa lhfV;ksa dh vko`fÙk 500 gVZ~t gS rFkk ok;q esa /ofu dh pky 340 eh@ls gS] O;fDr }kjk lquh x;h vko`fÙk;k¡ Kkr dhft;sA Ans. 529 Hz, 474 Hz 69.
A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4.0 beats per second. The speed of sound in air = 340 m/s. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train? /kheh xfr ls py jgh Vªsu ij lokj ,d ok;fyu oknd 440 gVZ~t dk ukn mRiUu dj jgk gSA jsy iFk ds lehi [kM+k
gqvk ,d vU; ok;fyu oknd leku ukn ct jgk gSA tc nksuksa lehi vk jgs gSa rFkk Vªsu tehu ij [kM+s ,d O;fDr ds lehi vk jgh gS] og 4.0 foLian izfr lsd.M lqurk gSA ok;q esa /ofu dh pky = 340 eh@lsA (a) Vªsu dh pky dh x.kuk dhft;sA (b) Vªsu esa fLFkr oknd fdruh foLian vko`fÙk lqusxk\ Ans. (a) 11 km/h 70.
(b) a little less than 4 beats/s
Two identical tuning fork vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0 m/s so that he approaches one tuning fork and recedes from the other (figure). Find the beat frequency observed by the listener. Speed of sound in air = 332 m/s. dqN nwjh ij fLFkr nks ,d tSls Lofj=k f}Hkqt leku vko`fÙk 256 gVZ~t ls dEiUu dj jgs gSaA nksuksa Lofj=k f}Hkqtksa ds e/; ,d Jksrk 3.0 eh@ls pky ls bl izdkj Hkkx jgk gS fd og ,d Lofj=k ds lehi ,oa nwljs ls nwj tk jgk gS ¼fp=k½A Jksrk }kjk lquh x;h foLian vko`fÙk Kkr dhft;sA ok;q esa /ofu dh pky = 332 eh@ls
Ans. 4.6 Hz 71.
Figure shows a person standing somewhere in between two identical tuning forks, each vibrating at 512 Hz. If both the tuning fork move towards right at a speed of 5.5 m/s, find the number of beats heard by the listener. Speed of sound in air = 330 m/s.
tSlk fd fp=k esa n'kkZ;k x;k gS nks ,d leku Lofj=k f}Hkqtksa ds e/; fdlh LFkku ij ,d O;fDr [kM+k gqvk gS] izR;sd 512 gVZ~t vko`fÙk ls dEiUu dj jgk gSA ;fn nksuksa Lofj=k f}Hkqt nka;h vksj 5.5 eh@ls pky ls xfr'khy gks tkrs gSa] Jksrk }kjk lqus x;s foLianksa dh la[;k Kkr dhft;sA ok;q esa /ofu dh pky = 330 eh@ls
Ans. 17.5 Hz, may not be able to distinguish 72.
A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius 100/p cm at a constant angular speed of 5.0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air = 332 m/s. 500 gVZ~t vko`fÙk ls dfEir ,d Nksvk /ofu lzksr 5.0 pDdj izfr lsd.M dh fu;r dks.kh; pky ls 100/ f=kT;k okys
o`Ùkkdkj iFk ij ?kwe jgk gSA ,d Jksrk o`Ùk ds ry esa fLFkr gSA /ofu dh U;wure rFkk vf/kdre izsf{kr vko`fÙk;k¡ Kkr dhft;sA ok;q esa /ofu dh pky = 332 eh@lsA Ans. 485 Hz and 515 Hz manishkumarphysics.in
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Chapter # 16 Sound Waves 73. Two trains are travelling towards each other both at a speed of 90 km/h. If one of the trains sounds a whistle at 500 Hz, what will be the apparent frequency heard in the other train ? Speed of sound in air = 350 m/s. ,d nwljs dh vksj xfr'khy nks Vªsusa] izR;sd 90 fdeh@?kaVk dh pky ls 500 gVZ~t vko`fÙk gVZ~t vko`fÙk dh lhVh ctkrh gS] nwljh Vªsu esa lquh x;h vkHkklh vko`fÙk fdruh gS\ ok;q esa /ofu dh pky = 350 eh@ls Ans. 577 Hz 74.
A traffic policeman sound a whistle to stop a car-driver approaching towards him. The car-driver does not stop and takes the plea in court that because of the Doppler shift, the frequency of the whistle reaching him might have gone beyond the audible limit of 20 kHz and he did not hear it. Experiments showed that the whistle emits a sound with frequency close to 16 kHz. Assuming that the claim of the driver is true, how fast was he driving the car? Take the speed of sound in air to be 330 m/s. Is this speed partical with today’s technology?
,d VªkfQd iqfyleSu viuh vksj vkrh gqbZ dkj ds Mªkboj dks jksdus ds fy;s lhVh ctkrk gSA dkj Mªkboj dkj ugha jksdrk gS rFkk U;k;ky; esa rdZ nsrk gS fd MkWIyj izHkko ds dkj.k lhVh JO; lhek 20 fdyks gVZ~t ls ijs pyh x;h vr% og ugha lqu ldkA iz;ksxksa }kjk ;g irk pyk fd lhVh dh vko`fÙk 16 fdyks gVZ~t ds vkl&ikl gSA eku yhft;s fd Mªkboj dk rdZ lgh gS] og dkj fdruh rst pyk jgk Fkk\ ok;q esa /ofu dh pky 330 eh@ls eku yhft;sA D;k vkidh rduhd ds vuqlkj ;g xfr laHko gS\ Ans. 300 km/h 75.
A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has a dominant frequency of 800 Hz. What will be the apparent frequency heard by the driver in the front car? Speed of sound in air = 330 m/s. 108 fdeh@?kaVk pky ls xfr'khy ,d dkj ns[krh gS fd mlds vkxs ,d vU; dkj mlh fn'kk esa 72 fdeh@?kaVk ls tk jgh gSA izFke dkj gkWuZ ctkrh gS] ftldh eq[; vko`fÙk 800 gVZ~t gSA vkxs okyh dkj ds Mªkboj }kjk lquh x;h vkHkklh vko`fÙk fdruh gS\ ok;q esa /ofu dh pky = 330 eh@ls Ans. 827 Hz
76.
Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/ h and the other at 54 km/h relative to the water. The first submarines sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz. (a) At what frequency is this signal received by the second submarine ? (b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be 1500 m/s. 'kkar leqnz esa nks i.MqfCc;k¡ ,d nwljs dh vksj vk jgh gSA izFke i.MqCch ty ds lkis{k 36 fdeh@?kaVk dh pky ls ;k=kk dj jgh gS rFkk nwljh 54 fdeh@?kaVk pky lsA izFke i.MqCch 2000 gVZ~t vko`fÙk dk ,d /ofu ladsr Hkstrh gSA ¼ikuh esa /ofu rjaxsa lksukj Hkh dgykrh gS½ (a) nwljh iuMqCch bl ladsr dks fdruh vko`fÙk dk xzg.k djsxh\ (b) nwljh iuMwCch bl ladsr dks fdruh vko`fÙk dk xzg.k djsxh\ ikuh esa /ofu dh pky 1500 eh@ls Ans. (a) 2034 Hz (b) 2068 Hz
77.
A small source of sound oscillates in simple harmonic motion with an amplitude of 17 cm. A detector is placed along the line of motion of the source. The source emits a sound of frequency 800 Hz which travels at a speed of 340 m/s. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source. ,d NksVk /ofu lzksr] 17 lseh vk;ke ls ljy vkorZ xfr dj jgk gSA lzksr dh xfr&js[kk ds vuqfn'k ,d lalwpd j[kk gqvk gSA lzksr 800 gVZ~t vko`fÙk dh /ofu mRlftZr djrk gS] tks 340 eh@ls pky ls xeu djrh gSA ;fn lalwpd }kjk lalwfpr /ofu dk cS.M&varjky 8 gVZ~t gS] lzksr dk vkorZdky Kkr dhft;sA Ans. 0.63 s
78.
A boy riding on his bike is going towards east at a speed of 4 2 m/s. At a certain point he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of 334 m/s. A second boy stands on the ground 45° south of east from him. Find the frequency of the pulse as received by the second boy. ,d yM+dk viuh lkbfdy ij lokj gksdj iwoZ dh vksj 4 2 eh@ls pky ls tk jgk gSA fdlh fof'k"V fcUnq ij og 1650 gVZ~t vko`fÙk dk /ofu Lian mRlftZr djrk gS tks ok;q esa 334 eh@ls pky ls xeu djrk gSA tehu ij ,d nwljk yM+dk mlls 45° iwoZ ls nf{k.k dh vksj [kM+k gSA nwlj yM+ds }kjk xzg.k dh x;h Lian dh vko`fÙk Kkr dhft;sA Ans. 1670 Hz
79.
A sound source, fixed at the origin is continuously emitting sound at a frequency of 660 Hz. The sound travels in air at a speed of 330 m/s. A listener is moving along the line x= 336 m/s. A listener is moving along the line x = 336 m at a constant speed of 26 m/s. Find the frequency of the sound as observed by the listener manishkumarphysics.in
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when he is (a) at y = – 140 m, (b) at y = 0 and (c) at y = 140 m. ewy fcUnq ij fLFkj ,d /ofu lzksr fujUrj 660 gVZ~t vko`fÙk dh /ofu mRlftZr dj jgk gSA ok;q esa /ofu 330 eh@ls pky ls xeu djrh gSA Jksrk x= 336 eh js[kk ds vuqfn'k 26 eh@ls dh fu;r pky ls xfr'khy gSA Jksrk }kjk izsf{kr /ofu dh vko`fÙk Kkr dhft;sA tc og gS % (a) y = – 140 eh ij , (b) y = 0 ij rFkk (c) y = 140 eh ij Ans. (a) 680 Hz (b) 660 Hz (c) 640 Hz 80.
A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. Speed of sound in air is 340 m/s. What frequency will a passenger sitting near the ope window hear? (b) What frequency will a person standing near the track hear whom the train has just passed? (c) A wind starts blowing towards east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track. iwoZ dh vksj 108 fdeh@?kaVk pky ls xfr'khy ,d Vªsu 500 gVZ~t eq[; vko`fÙk dh lhVh ctkrh gSA (a) [kqyh gqbZ f[kM+dh ds ikl cSBk gqvk ;k=kh fdruh vko`fÙk lquxs k\ (b) jsy iFk ds ikl [kM+k gqvk O;fDr] tSlh gh mlds ikl ls Vªus xqtjsxh] rqjar fdruh vko`fÙk lqusxk\ (c) ok;q 36 fdeh@?kaVk dh pky ls iwoZ dh vksj izokfgr gksuk izkjEHk dj nsrh gSA Vªsu ;k=kh
rFkk jsy iFk ds lehi [kM+s O;fDr }kjk lquh x;h vko`fÙk;ksa dh x.kuk dhft;sA
Ans. (a) 500 Hz (b) 459 Hz (c) 500 Hz by the passanger and 458 by the person near the track 81.
A boy riding on a bicycle going at 12 km/h towards a vertical wall wishes at his dog on the ground. If the frequency of the whistle is 1600 Hz and the speed of sound in air is 330 m/s, find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy. ,d m/okZ/kj nhokj dh vksj 12 fdeh@?kaVk dh pky ls tkrk gqvk lkbfdy ij lokj yM+dk tehu ij [kM+s gq, mlds dqÙks dh vksj lhVh ctkrk gSAa ;fn lhVh dh vko`fÙk 1600 gV~t Z gS rFkk ok;q esa /ofu dh pky 330 eh@ls gS] Kkr dhft;s% (a) nhokj }kjk xzg.k dh x;h /ofu dh vko`fÙk (b) yM+ds }kjk xzg.k dh x;h lhVh dh ijkofrZr gksdj vkus okyh vko`fÙkA Ans.(a) 1616 Hz (b) 1632 Hz
82.
A person standing on a road sends a sound signal to the driver of a car going away from him at a speed of 72 km/h. The signal travelling at 330 m/s in air and having a frequency of 1600 Hz gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person. lM+d ij [kM+k gqvk ,d O;fDr] mlls 72 fdeh@?kaVk dh pky ls nwj tkrh gqbZ ,d dkj dh vksj /ofu ladsr Hkstrk gSA /ofu ladsr ok;q esa 330 eh@ls pky ls xeu djrk gS rFkk bldh vko`fÙk 1600 gVZ~t gS] ;g dkj dh ckWMh ls
ijkofrZr gksdj iqu% ykSVrk gSA O;fDr }kjk lquh x;h ijkofrZr ladsr dh vko`fÙk Kkr dhft;sA Ans. 1417 Hz 83.
A car moves with a speed of 54 km/h towards a cliff. The horn of the car emits sound of frequency 400 Hz at a speed f 335 m/s. (a) Find the wavelength of the sound emitted by the horn in front of the car. (b) Find the wavelength of the wave reflected from the cliff. (c) What frequency does a person sitting in the car hear for the reflected sound wave? (d) How many beats does he hear in 10 seconds between the sound coming directly from the horn and that coming after the reflection? ,d dkj 54 fdeh@?kaVk pky ls ,d Å¡ph pV~Vku dh vksj xfr'khy gSA dkj dk gkWuZ 400 gVZ~t vko`fÙk dh /ofu dh mRlftZr djrk gSA ftldh pky 335 eh@ls gSA (a) dkj ds lkeus mRlftZr /ofu dh rjaxnS/;Z Kkr dhft;sA (b) pV~Vku ls ijkofrZr rjax dh rjaxnS/;Z Kkr dhft;sA (c) dkj esa cSBk gqvk O;fDr ijkofrZr /ofu rjax dh fdruh vko`fÙk lqusxk\ (d) og gkuZ ls lh/kh vkus okyh /ofu rFkk ijkorZu ds i'pkr~ vkus okyh /ofu rFkk ijkorZu ds i'pkr~ vkus okyh /ofu esa 10 lsd.M esa fdrus foLian lqusxk\ Ans. (a) 80 cm (b) 80 cm (c) 437 hz (d) No beat may be heard
84.
An operator sitting in his base camp sends a sound signal of frequency 400 hz. The signal is reflected back from a car moving towards him. The frequency of the reflected sound is found to be 410 hz. Find the speed of the car. Speed of sound in air = 324 m/s. vius vk/kkj dsEi ij cSBk gqvk ,d vkWijsVj 400 gVZ~t vko`fÙk dk /ofu ladsr Hkstrk gSA mldh vksj vkrh gqbZ dkj ls ;g ladsr ijkofrZr gksdj okfil ykSVrk gSA ijkofrZr gksdj okfil ykSVrk gSA ijkofrZr /ofu dh vko`fÙk 410 gV~Zt izsf{kr gksrh gSA dkj dh pky Kkr dhft;sA ok;q esa /ofu dh pky = 324 eh@lsA Ans. 12 m/s
85.
Figure shows a source of sound moving along the X-axis at a speed of 22 m/s continuously emitting a sound of frequency 2.0 kHz which travels in air at a speed of 330 m from the origin. At t = 0, the source crosses the origin P. (a) When does the sound emitted from the source at P reach the listener Q? (b) What will be the frequency heard by the listener at this instant? (c) Where will the source be at this instant ? fp=k esa iznf'kZr fd;k x;k gS fd X-v{k ds vuqfn'k 22 eh@ls- pky ls xfr'khy ,d /ofu L=kksr fujUrj 2.0 fdyks gV~t Z manishkumarphysics.in
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vko`fÙk dh /ofu mRlftZr dj jgk gS] tks ok;q esa 330 eh-@ls pky ls xeu dj jgh gSA Y v{k ij ewy fcUnq P ls 330 eh- nwj ,d Jksrk Q [kM+k gqvk gSA t = 0 ij L=kksr ewy fcUnq P dks ikj djrk gSA
Ans. (a) t = 1 sec
(b) 2.0 k Hz
(c) at x = 22 metre
86.
A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22m/s. A listener is situated on the ground at the point (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. Speed of sound in air = 330m/s. 4000 gVZ~t vko`fÙk dh /ofu mRlftZr dj jgk ,d lzksr Y-v{k ds vuqfn'k 22 eh@ls pky ls xfr'khy gSA ,d Jksrk tehu ij ¼660 eh, 0) fLFkfr ij fLFkr gSA ftl {k.k ij lzksr ewy fcUnq dks ikj djrk gS] Jksrk }kjk xzg.k dh x;h /ofu dh vko`fÙk Kkr dhft;sA ok;q esa /ofu dh pky = 330 eh@lsA Ans. 4018 Hz
87.
A source of sound emitting a 1200 Hz note travels along a straight line at a speed of 170 m/s. A detector is placed at a distance of 200 m from the line of motion of the source. (a) Find the frequency of sound received by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant is detects the frequency 1200 Hz. Velocity of sound in air = 340 m/s. ,d /ofu lzksr 1200 gVZ~t vko`fÙk dk ukn mRiUu djrs gq, ,d ljy js[kk ds vuqfn'k 170 eh@ls pky ls xfr'khy gSA lzksr dh xfr dh js[kk ls 200 eh nwjh ij ,d lalwpd j[kk gqvk gSA (a) ftl {k.k ij lzksr fudVre gSA lalwpd }kjk xzg.k dh x;h /ofu dh vko`fÙk Kkr dhft;s (b) ftl {k.k ij lalwpd 1200 gVZ~t vko`fÙk lalwpd djrk gS] lzksr rFkk lalwpd ds e/; nwjh Kkr dhft;sA ok;q esa /ofu dh pky = 340 eh@ls Ans. (a) 1600 Hz (b) 224 m
88.
A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point. (a) An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle (figure). The speed of sound in air = 330 m/s and g 10 m/s2. Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the centre of the circle. Find the frequencies heard by the observer corresponding to the sound emitting by the source when it is at the same height as the centre. 500 gVZ~t vko`fÙk dh /ofu mRlftZr dj jgk ,d NksVk /ofu lzksr] ,d gYdh Mksjh ls cka/kdj 1.6 eh f=kT;k okys m/ oZ o`Ùk esa ?kqek;k tk jgk gSA tc lzksr mPpre fcUnq ij gS] Mksjh ek=k ruko ;qDr jgrh gSA (a) ,d izs{kd blh m/oZ o`Ùk ds ry esa rFkk o`Ùk ds dsUnz ds leku Å¡pkbZ ij cgqr nwj fLFkr gSA ok;q esa /ofu dh pky = 330 eh@ls rFkk g = 10 eh@ls2 gSA izs{kd }kjk lqu h x;h vf/kdre vko`fÙk Kkr dhft;sA (b) ,d izs{kd o`Ùk ds dsUnz ls m/okZ/kj Åij
cgqr vf/kd nwjh ij gSA tc lzksr o`Ùk ds dsUnz ds cjkcj Å¡pkbZ ij gSA izs{kd }kjk lquh gqbZ vko`fÙk Kkr dhft;sA
Ans. (a) 506 Hz 89.
(b) 490 Hz and 511 Hz
A source emitting a sound of frequency v is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v. v vko`fÙk dh /ofu mRlftZr dj jgk ,d lzksr ,d izs{kd ls cgqr vf/kd nwjh ij j[kk gqvk gSA lzksr fu;r Roj.k ls
izs{kd dh xfr djuk izkjEHk djrk gSA lzksr ds xfr izkjEHk djrs gh mRlftZr rjax ds laxr izs{kd }kjk lquh x;h vko`fÙk Kkr dhft;sa ek/;e esa /ofu dh pky v gSA Ans.
2uv 2 2uv a
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