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Waves MCQs for NEET — Physics Questions with Answers

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The Doppler's effect is applicable for :

Light waves

Sound waves

Space waves

Both (1) and (2)

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Explanation

Doppler’s effect is applicable for both light and sound waves.

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A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

9 : 8

8 : 9

1 : 1

9 : 10

(Speed of sound v = 340 m/s)

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Explanation

When source is approaching the observer, the frequency heard

$n_{a} = (\frac{v}{v - v_{S}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$

When source is receding, the frequency heard $n_{r} = (\frac{v}{v + v_{S}}) \times n$ = $\frac{340}{340 + 20} \times 1000 = 944$

$\Rightarrow n_{a} : n_{r} = 9 : 8$

Short tricks : $\frac{n_{a}}{n_{r}} = \frac{v + v_{S}}{v - v_{S}} = \frac{340 + 20}{340 - 20} = \frac{9}{8} .$

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A source and listener are both moving towards each other with speed v/10, where v is the speed of sound. If the frequency of the note emitted by the source is f, the frequency heard by the listener would be nearly :

1.11 f

1.22 f

1.27 f

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Explanation

When the source and listener are moving towards each other, the apparent frequency heard by the listener is higher due to the Doppler effect. If they are moving with speed v/10 (where v is the speed of sound), the apparent frequency is increased by a factor of (1 + v/10)/(1 - v/10), which is approximately 1.22.

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A motor car blowing a horn of frequency 124vib/sec moves with a velocity 72 km/hr towards a tall wall. The frequency of the reflected sound heard by the driver will be (velocity of sound in air is 330 m/s)

109 vib/sec

132 vib/sec

140 vib/sec

248 vib/sec

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A source of sound of frequency n is moving towards a stationary observer with a speed S. If the speed of sound in air is V and the frequency heard by the observer is n₁, the value of n₁/n is

(V + S)/V

V/(V + S)

(V – S)/V

V/(V – S)

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Explanation

By using $n^{'} = n \frac{v}{v - v_{S}}$ ⇒ $\frac{n_{1}}{n} = (\frac{V}{V - S})$

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An observer is moving away from the source of the sound of frequency 100 Hz. His speed is 33 m/s. If the speed of sound is 330 m/s, then the observed frequency is :

90 Hz

100 Hz

91 Hz

110 Hz

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Explanation

$n' = n (\frac{v - v_{O}}{v}) = (\frac{330 - 33}{330}) \times 100 = 90 H z$

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An observer standing at the station observes frequency 219 Hz when a train approaches and 184 Hz when train goes away from him. If velocity of sound in air is 340 m/s, then velocity of train and actual frequency of whistle will be

$15 .5 m s^{- 1}, 200 H z$

$19 .5 m s^{- 1}, 205 H z$

$29 .5 m s^{- 1}, 200 H z$

$32 .5 m s^{- 1}, 205 H z$

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Explanation

When train is approaching frequency heard by the observer is

$n_{a} = n (\frac{v}{v - v_{S}})$ ⇒ $219 = n (\frac{340}{340 - v_{S}})$ …(i)

when train is receding (goes away), frequency heard by the observer is

$n_{r} = n (\frac{v}{v + v_{s}})$ ⇒ $184 = n (\frac{340}{340 + v_{s}})$ …(ii)

On solving equation (i) and (ii) we get

$n = 200 H z$ and $v_{S} = 29 .5 m / s .$

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A boy is walking away from a wall towards an observer at a speed of 1 metre/sec and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observer per second is (Velocity of sound in air = 340 metres/sec

Zero

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Explanation

Observer hears two frequencies

(i) n₁ which is coming from the source directly

(ii) n₂ which is coming from the reflection image of source

so, $n_{1} = 680 (\frac{340}{340 - 1})$ and $n_{2} = 680 (\frac{340}{340 + 1})$

$\Rightarrow n_{1} - n_{2} = 4$ beats

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The driver of a car travelling with speed 30 metres per second towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 metres per second, the frequency of the reflected sound as heard by the driver is :

720 Hz

555.5 Hz

550 Hz

500 Hz

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Explanation

From the figure, it is clear that

Frequency of reflected sound heard by the driver.

$n' = n [\frac{v - (- v_{o})}{v - v_{s}}] = n [\frac{v + v_{o}}{v - v_{s}}] = n [\frac{v + v_{c a r}}{v - v_{c a r}}]$

$= 600 [\frac{330 + 30}{330 - 30}] = 720 H z .$

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A source of sound is travelling with a velocity 40 km/hour towards the observer and emits the sound of frequency 2000 Hz. If the velocity of sound is 1220 km/hour, then what is the apparent frequency heard by an observer

2210 Hz

1920 Hz

2068 Hz

2086 Hz

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Explanation

$n^{'} = n (\frac{v}{v - v_{S}}) = \frac{2000 \times 1220}{(1220 - 40)} = 2068 H z$

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