Questions on Doppler Effect in sound often appear in Medical and Engineering Entrance tests. The general expression for the apparent frequency (n') is

It will be helpful to remember that if the source moves towards the listener or the listener moves towards the source, the apparent frequency increases. If they move away, the apparent frequency decreases.

You may find questions on Doppler Effect in light occasionally. As light and other electromagnetic radiations do not require a medium for their transmission, the velocity (w) of the medium will be absent in the expression for the apparent frequency. Instead of the speed of sound ‘v’ you will have the constant speed ‘c’ of light. The expression for the apparent frequency is

If a star moves towards the observer, as in the case of a member of a binary stars system, the shift is ‘

Since the speed of light is much greater than the speed of material bodies, the Doppler shift for a given relative velocity is the same, whether the source moves or the observer moves.

Now consider the following MCQ which appeared in Kerala Engineering Entrance 2002 test paper:

In the expression for the apparent frequency of the reflected sound, the wind velocity (w) is zero. The velocity of source (v

Number of beats per second = 280-272 = 8.

The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:

Now consider the following question:

The following MCQ appeared E.A.M.C.E.T. (Med.) A.P. 2003 question paper:

v

The velocity of aircraft is half that of the reflected image (source). So the answer is 50 m/s.

Now consider the following question:

Note that you can substitute the wave length in the given unit itself since z is a ratio of wave lengths.

**n' = n(v+w-v**where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘v_{L})/(v+w-v_{S})_{L}’ is the velocity of the listener, ‘v_{S}’ is the velocity of the source of sound and ‘n’ is the real frequency of the sound emitted by the source. You should clearly note that all the velocities in this expression are in the same direction and the source is behind the observer.It will be helpful to remember that if the source moves towards the listener or the listener moves towards the source, the apparent frequency increases. If they move away, the apparent frequency decreases.

You may find questions on Doppler Effect in light occasionally. As light and other electromagnetic radiations do not require a medium for their transmission, the velocity (w) of the medium will be absent in the expression for the apparent frequency. Instead of the speed of sound ‘v’ you will have the constant speed ‘c’ of light. The expression for the apparent frequency is

**n' = n(c-v**

Doppler Effect in light is used to calculate the recessional velocities of stars and galaxies by measuring the ‘_{L})/(c-v_{S})**. When the star moves away from the observer, as is the case with our expanding universe, the wave length of the light emitted by the star appears to be increased (shifted towards red colour). From the above expression for apparent frequency, the shift in frequency which is equal to (n-n') is related to the recessional velocity v***red shift’*_{S}by the equation.**v**

The above expression can be written in terms of the shift in wave length as_{S}= c( n-n' )/n'**v**

This relation is often written as_{S}= c(λ'-λ)/λ**v = zc**where**z =(λ'-λ)/λ**which may be called the ‘spectral shift’ or ‘Doppler shift’.If a star moves towards the observer, as in the case of a member of a binary stars system, the shift is ‘

**and the two expressions for the velocity of the star are***blue shift’***v**and_{S}= c(n'-n)/n'**v**

You may get questions involving the calculation of the recessional velocities of stars using the red shift. Questions involving blue shift as well as red shift may be asked for calculating the velocity of aircraft and automobiles which reflect microwaves transmitted by a radar._{S}= c( λ-λ' )/λSince the speed of light is much greater than the speed of material bodies, the Doppler shift for a given relative velocity is the same, whether the source moves or the observer moves.

Now consider the following MCQ which appeared in Kerala Engineering Entrance 2002 test paper:

**A person carrying a whistle, emitting continuously a note of frequency 272Hz is running towards a reflecting surface with a speed of 18km/hour. The speed of sound in air is 345m/s. The number of beats heard by him per second is**

(a) 4 (b) 6 (c) 8 (d) 3 (e) zero

Beats are produced here because of the superposition of the direct sound of actual frequency n (equal to 272Hz) from the whistle and the reflected sound of apparent frequency n’, given by n’ = n(v+w-v(a) 4 (b) 6 (c) 8 (d) 3 (e) zero

_{L})/(v+w-v_{S}). [Note that The frequency of the direct sound is unchanged since both the source and listener are moving together and there is no relative velocity between them].In the expression for the apparent frequency of the reflected sound, the wind velocity (w) is zero. The velocity of source (v

_{S}) is positive since the source is the reflected sound image of the whistle and this image moves towards the listener. The velocity of the listener is negative since he is moving towards the reflected image serving as the source. Therefore, n' = n(v+v_{L})/(v-v_{S}) = 272(345+5)/(345-5) =280Hz. [Note that 18km/hour = 5m/s].Number of beats per second = 280-272 = 8.

The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:

**The apparent frequency of the whistle of an engine changes in the ratio 9:8 as the engine passes a stationary observer. If the velocity of the sound is 340ms**

(a) 40 m/s (b) 20 m/s (c) 340 m/s (d) 180 m/s (e) 50 m/s

Problems of this type are often found in entrance test papers. When the engine approaches the observer, the apparent frequency is v/(v-v^{-1}, then the velocity of the engine is(a) 40 m/s (b) 20 m/s (c) 340 m/s (d) 180 m/s (e) 50 m/s

_{S}) and when it moves away from the observer, the apparent frequency is v/(v+v_{S}). The frequency therefore changes in the ratio (v+v_{S})/(v-v_{S}). We have therefore, (v+v_{S})/(v-v_{S}) =9/8 from which the velocity of the source, v_{S}= v/17 = 340/17 = 20m/s.Now consider the following question:

**Two trains A and B approach a station from opposite sides, sounding their whistles. A stationary observer on the platform hears no beats. If the velocities of A and B are 15m/s and 30m/s respectively and the real frequency of the whistle of B is 600 Hz, the real frequency of the whistle of A is ( Velocity of sound = 330m/s)**

(a) 660 Hz (b) 630 Hz (c) 600 Hz (d) 570 Hz (e) 540 Hz

This is a case increase in apparent frequency due to the motion of the source. Since there are no beats, the apparent frequency [v/(v-v(a) 660 Hz (b) 630 Hz (c) 600 Hz (d) 570 Hz (e) 540 Hz

_{S})]n of the whistle of A is the same as that of B so that we have, [330/(330-15)]n = [330/330-30)]×600 so that n/315 = 600/300 from which n = 630 Hz.The following MCQ appeared E.A.M.C.E.T. (Med.) A.P. 2003 question paper:

**A radar sends a radio signal of frequency 9×10**

(a) 150 (b) 100 (c) 50 (d) 25

The source of waves in this problem is the reflected image of the radar, the aircraft functioning as a mirror. The velocity of the reflected image is twice the velocity of the ‘mirror’(aircraft).You have to note this while calculating the velocity (v^{9}Hz towards an aircraft approaching the radar. If the reflected wave shows a frequency shift of 3×10^{3}Hz, the speed with which the aircraft is approaching the radar in m/s is (velocity of radio signal is 3×10^{8}m/s)(a) 150 (b) 100 (c) 50 (d) 25

_{S}) of the source using the equation involving ‘blue shift’ discussed in the beginning of this post:v

_{S}= c(n'-n)/n' = 3×10^{8}(3×10^{3})/(9×10^{9}) = 100 m/s.The velocity of aircraft is half that of the reflected image (source). So the answer is 50 m/s.

Now consider the following question:

**A spectral line of wave length 5000 A.U. in the light coming from a distant star is observed to be shifted to 5004 A.U. The component of the recessional velocity of the star along the line of sight is**

(a) 1.4×10

We have v = zc where z = (λ'-λ)/λ. Therefore, the recessional velocity = [(5004-5000)/5000] ×3×10(a) 1.4×10

^{5}m/s (b) 1.8×10^{5}m/s (c) 2.0×10^{5}m/s (d) 2.4×10^{5}m/s (e) 1.4×10^{4}m/s^{8}= (4×3×10^{8})/5000 = 2.4×10^{5}m/s.Note that you can substitute the wave length in the given unit itself since z is a ratio of wave lengths.

The following MCQ appeared in IIT-JEE 2003 question paper:

n(v-v

176(330-v

**A police car moving at 22 m/s, chases a motor cyclist. 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 motor cyclist, if it is given that he does not hear any beats. (Velocity of sound = 330 m/s)**

(a) 33 m/s (b) 22 m/s (c) zero (d) 11 m/s

Since the motor cyclist does not hear any beats, the apparent frequency of the horn of the police car (as heard by the motor cyclist) is the same as the apparent frequency of the stationary siren. Therefore, we have,(a) 33 m/s (b) 22 m/s (c) zero (d) 11 m/s

n(v-v

_{L})/(v-v_{S}) = n'(v+v_{L})/v where v is the velocity of sound, v_{L}is the velocity of the motor cyclist, v_{S}is the velocity of the police car and n and n' are the frequencies of the police car horn and the stationary siren respectively. Substituting for the velocities, we obtain176(330-v

_{L})/(330-22) = 165(330+v_{L})/330, from which v_{L}= 22 m/s.
sir,

ReplyDeleteFor sign convention for Vs, VL. I will arrange the general eqn i.e.,

n(v-vl)/(v-vs) such that,

if listener or source is approaching towards the other, the apparent freq should be more or else the apparent frequency is less.

Am I correct on this?

Yes, you are correct.

ReplyDelete