IIT-JEE 2011 and IIT-JEE 2010 Questions on Doppler Effect

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."
– Sir Isaac Newton

Questions on Doppler Effect were discussed on this site earlier. You can access them by clicking on the label ‘Doppler effect’ below this post or by trying a search for ‘Doppler effect’ using the search box provided on this page. Questions on Doppler effect appear frequently in Entrance examination question papers. Today we will discuss two questions in this section Here is the first question which appeared in IIT-JEE 2011 question paper as a single answer type multiple choice question:

(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

(A) 8.50 kHz

(B) 8.25 kHz

(C) 7.75 kHz

(D) 7.50 kH

The general expression for the apparent frequency (n’) produced due to Doppler effect is

n’ = n(v+w–v_L)/(v+w–v_S) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘v_L’ is the velocity of the listener, ‘v_S’ is the velocity of the source of sound and ‘n’ is the actual frequency of the sound emitted by the source. Note that all the velocities in this expression are in the same direction and the source is behind the listener. In other words, the listener is moving away from the source and the source is moving towards the listener.

[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].

In the present case the source of sound which produces the Doppler effect is the reflected image of the siren. (The car driver will not detect any change in the frequency of the direct sound from the siren since he is moving along with the siren). The reflected image (source) of the siren moves towards the car driver with a speed of 36 km/hr and the car driver (listener) moves towards the reflected image (source) with the same speed.

Thus in the expression for apparent frequency we have ro substitute

n = 8 kHz, v_L = – 36 km/hr = 36×(5/18) ms^–1 = –10 ms^–1, w = 0, v_S = + 10 ms^–1.

Therefore, n’ = 8×(320 + 10)/(320 – 10) = 8×(33/31) kHz = 8.5 kHz.

The following question appeared in IIT-JEE 2010 question paper as an integer type question (in which the answer is a single-digit integer, ranging from 0 to 9):

(2) 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.

This is similar to question no.1 above. The difference between the frquencies of the sound reflected from the cars is given by

n₁’ – n₂’ = [f₀(v+v₁)/(v–v₁)] – [f₀(v+v₂)/(v–v₂)] where v₁ and v₂ are the speeds of the two cars.

Therefore (n₁’ – n₂’)/f₀ = [(v+v₁)/(v–v₁)] – [(v+v₂)/(v–v₂)]

The quantity on the left hand side of the above equation is 1.2/100 as given in the question.

Therefore, 0.012 = [(v+v₁) (v–v₂) – (v+v₂) (v–v₁)] / [(v–v₁)(v–v₂)]

Or, 0.012 = (2 vv₁ – 2 vv₂)/ [(v–v₁)(v–v₂)] = 2v(v₁ –v₂)/v² since (v–v₁) ≈ (v–v₂) ≈ v

[The cars are moving at speeds much smaller than the speed of sound].

Therefore we have

0.012 = 2(v₁ –v₂)/v from which (v₁ –v₂) = 0.006×330 ms^–1

The difference in the speeds of the cars in km per hour is 0.006×330×(18/5) = 7.128

The answer, which is the nearest integer, is 7.

Two Questions (MCQ) on Doppler Effect in Sound

You will find the formulae to be remembered in connection with Doppler effect and some useful multiple choice questions (with solution) on Doppler effect at this location on this site.

[If you want all questions on Doppler effect on this site, you may click on the label ‘Doppler effect’ below this post].

Today I give you two more multiple choice questions on Doppler effect:

(1) A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms^–1. The velocity of sound in air is 300 ms^–1. If the person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the whistle is

(1) 30 ms^–1

(2) 15√2 ms^–1

(3) 15/√2 ms^–1

(4) 15 ms^–1

The apparent frequency (n’) in terms of all possible variables is given by

n’ = n(v+w–v_L)/(v+w–v_S) where ‘n’ is the real frequency of sound, ‘v’ is the velocity of sound, ‘w’ is the velocity of wind, ‘v_L’ is the velocity of listener and ‘v_S’ is the velocity of the source of sound. Note that in the above expression all velocities are in the same direction and the source is behind the listener and is therefore approaching the listener.

Since the wind velocity is zero and the listener is stationary in the first case, the above expression reduces to

n’ = nv/(v–v_S)

[The apparent frequency is therefore greater than the real frequency].

As the person can hear frequencies up to 10000 Hz only, the maximum value of v_S upto which he can hear the whistle is given by

10000 = 9500×300/(300 – v_S)

Therefore, 300 – v_S = 285 so that v_S = 15 ms^–1

[The above question appeared in AIEEE 2006 question paper]

(2) A person moves away with constant velocity ‘v₀’ from a stationary train which blows its whistle. The ratio of the real frequency of the whistle to the apparent frequency as measured by the person is 1.25. If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the ratio of the real frequency of the whistle to the apparent frequency as measured by the person will be

(a) 1.2

(b) 1.25

(c) 1.4

(d) 1.45

(e) 1.5

The apparent frequency (n₁) as measured by the person moving away from the whistle with velocity v_L is given by

n₁ = n(v–v_L)/v where ‘n’ is the real frequency of the whistle and ‘v’ is the velocity of sound.

Therefore, n/n₁ = v/(v–v_L) = v/(v–v₀).

Since n/n₁ = 1.25,we have v/(v–v₀) =1.25 so that 1– (v₀/v) = 0.8 from which v₀ = 0.2v.

If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the apparent frequency (n₂) as measured by the person is given by

n₂ = nv/(v +v_S) where v_S = v₀

Therefore, n/n₂ = (v+v₀)/v.

Substituting for v₀ (= 0.2v) we obtain n/n₂ = (v+0.2v)/v = 1.2

Multiple Choice Questions on Doppler Effect

Questions on Doppler Effect in sound often appear in Medical and Engineering Entrance tests. The general expression for the apparent frequency (n') is
n' = n(v+w-v_L)/(v+w-v_S) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘v_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_L)/(c-v_S)
Doppler Effect in light is used to calculate the recessional velocities of stars and galaxies by measuring the ‘red shift’. 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_S by the equation.
v_S = c( n-n' )/n'
The above expression can be written in terms of the shift in wave length as
v_S = c(λ'-λ)/λ
This relation is often written as 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 ‘blue shift’ and the two expressions for the velocity of the star are
v_S = c(n'-n)/n' and v_S = c( λ-λ' )/λ
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.
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:
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_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^-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
Problems of this type are often found in entrance test papers. When the engine approaches the observer, the apparent frequency is v/(v-v_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_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 9×10⁹ Hz towards an aircraft approaching the radar. If the reflected wave shows a frequency shift of 3×10³ Hz, the speed with which the aircraft is approaching the radar in m/s is (velocity of radio signal is 3×10⁸ m/s)
(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_S) of the source using the equation involving ‘blue shift’ discussed in the beginning of this post:
v_S = c(n'-n)/n' = 3×10⁸(3×10³)/(9×10⁹) = 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) 1.4×10⁵ m/s (b) 1.8×10⁵ m/s (c) 2.0×10⁵ m/s (d) 2.4×10⁵ m/s (e) 1.4×10⁴ m/s
We have v = zc where z = (λ'-λ)/λ. Therefore, the recessional velocity = [(5004-5000)/5000] ×3×10⁸ = (4×3×10⁸)/5000 = 2.4×10⁵ 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:
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,
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 obtain
176(330-v_L)/(330-22) = 165(330+v_L)/330, from which v_L = 22 m/s.