Q 1 :

A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is                [2011]

  • 8.50 kHz

     

  • 8.25 kHz

     

  • 7.75 kHz

     

  • 7.50 kHz

     

(1)

 



Q 2 :

A police car moving at 22 m/s, chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that he does not observes any beats.                     [2003]

  • 33 m/s

     

  • 22 m/s

     

  • zero

     

  • 11 m/s

     

(2)

f1=frequency of the police car heard by motorcyclist,

f2=frequency of the siren heard by motorcyclist.

f1=330-v330-22×176;    f2=330+v330×165;

  f1-f2=0    v=22 m/s



Q 3 :

A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that train A is     [2002]

  • 242252

     

  • 2

     

  • 56

     

  • 116

     

(2)

Using the formula  f'=f(vA+vv)

vA+vv=5.55     and     VB+VV=65

  vBvA=2



Q 4 :

A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed v with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed v with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be n Hz. The value of n is _____.                 [2024]



(200)

 Using Doppler's effect, frequency received by observer

f0=(C±V0C±Vs)fs,    C=speed of sound

Case-1:  f1=(C+VC-V)fs

  288=(C+VC-V)240               .....(i)

Case-2:  f2=(C-VC+V)fs

or,  n=(C-VC+V)240               .....(ii)

Multiplying eq. (i) & (ii)

(288)(n)=(240)(240)

  n=240×240288=200



Q 5 :

A train S1, moving with a uniform velocity of 108 km/h approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36 km/h towards S2 as shown in figure. Both the trains are blowing whistles of same frequency 120 Hz. When O is 600 m away from S2 and distance between S1 and S2 is 800 m, the number of beats heard by O is _____

[Speed of the sound = 330 m/s]                    [2019]



(8.13)

 Apparent frequency heard by observer due to source S1

ν1=ν[v+v0v]=120[330+10330]=120×3433=123.636 Hz

Apparent frequency heard by observer due to source S2

ν2=ν[v+v0v-vs]=120[330+10cos53°330-30cos37°]

  ν2=120[330+10×0.6330-30×0.8]=120[336306]=131.764 Hz

  Beat frequency, νb=ν2-ν1=131.764-123.636=8.125 Hz8.13 Hz



Q 6 :

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 ms-1 and the man behind walks at a speed 2.0 ms-1. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 ms-1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is ____________.                 [2018]



(5)

 Apparent frequency at O due to source at -A

νA=ν[vv-2cosθ]

Apparent frequency at O due to source at -B

νB=ν[vv+cosθ]

  Beat frequency, νb=ν[vv-2cosθ]-ν[vv+cosθ]

=νv[1v-2cosθ-1v+cosθ]

=1430×330[1330-2×513-1330+513]

=1430×330×13[1330×13-10-1330×13+5]

=1430×330×13[14280-14295]5 Hz



Q 7 :

A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms-1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz?

(Given that the speed of sound in air is 330 ms-1 and the car reflects the sound at the frequency it has received.)                [2017]



(6)

 Frequency observed at car

ν1=ν0(v+vcv)

Frequency of reflected sound as observed at the source

ν2=ν1(vv-vc)=ν0(v+vcv-vc)

  Beat frequency=ν2-ν0

=ν0[v+vc-v+vcv-vc]=ν0[2vcv-vc]

=492[2×2330-2]=492×4328=6 Hz



Q 8 :

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π3, 2π3 and π. When they are superposed, the intensity of the resulting wave is nI0. The value of n is                            [2015]



(3)

 Resultant of amplitude, AR

=I0[sin0+sinπ3+sin2π3+sinπ]

=I0[32+32]=3I0

  IR=AR2=3I0        n=3



Q 9 :

A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms-1.                    [2010]



(7)

Firstly, car will be treated as an observer which is approaching the source. Then it will be treated as a source, which is moving in the direction of sound.

Frequency of sound reflected by the car C1

f1=f0(v+v1v-v1)

And frequency of sound reflected by the car C2

f2=f0(v+v2v-v2)

  f1-f2=(1.2100)f0=f0[v+v1v-v1-v+v2v-v2]

or  (1.2100)f0=2v(v1-v2)(v-v1)(v-v2)f0

(v-v1)=(v-v2)v  as v1 and v2 are very very less than v.

  (1.2100)f0=2(v1-v2)vf0

or  (v1-v2)=v-1.2200=330×1.2200=1.98 m s-17 km h-1



Q 10 :

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 ms-1 in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320 ms-1, the smallest value of the percentage change required in the length of the pipe is __________.                      [2020]



(0.62 to 0.63)

 Let 1=initial length of pipe

2=new length of pipe

VT=Speed of tuning fork

In closed organ pipe,    f=V41

When tuning fork is moved,    f'=f(VV-VT)=V42

V41(VV-VT)=V42V-VTV=21

21-1=V-VTV-12-11=-VTV

Percentage change required in the length of the pipe

2-11×100=-2320×100=-0.625%

Hence, smallest value of percentage change required in the length of pipe is 0.625%



Q 11 :

S1 and S2 are two identical sound sources of frequency 656 Hz. The source S1 is located at O and S2 moves anticlockwise with a uniform speed 42 ms-1 on a circular path around O, as shown in the figure. There are three points P, Q and R on this path such that P and R are diametrically opposite while Q is equidistant from them. A sound detector is placed at point P. The source S1 can move direction OP.

[Given: The speed of sound in air is 324 ms-1]

Q.   When only S2 is emitting sound and it is at Q, the frequency of sound measured by the detector in Hz is __________.            [2023]



(648)

 Given: velocity of sound V=324 m/s

Frequency of source f0=656 Hz

Velocity of source S2,    Vs=42cos45°

From Doppler's effect, apparent frequency measured by the detector, f=(VV+Vs)f0

=324324+42×12×656

=648 Hz



Q 12 :

S1 and S2 are two identical sound sources of frequency 656 Hz. The source S1 is located at O and S2 moves anticlockwise with a uniform speed 42 ms-1 on a circular path around O, as shown in the figure. There are three points P, Q and R on this path such that P and R are diametrically opposite while Q is equidistant from them. A sound detector is placed at point P. The source S1 can move direction OP.

[Given: The speed of sound in air is 324 ms-1]

Q.   Consider both sources emitting sound. When S2 is at R and S1 approaches the detector with a speed 4 ms-1, the beat frequency measured by the detector is __________ Hz.                    [2023]



(8.2)

f0=656 Hz

v=324 m/s

Frequency heard due to movement of (S1)

f1=(vv-us)f0

f1=324320×656

And frequency heard due to movement of (S2)

f2=656 Hz

  Beat frequency Δf=f1-f2

=656(324320-1)

Δf=8.2