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Waves MCQs for NEET — Physics Questions with Answers

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A source of sound and listener are approaching each other with a speed of 40 m/s. The apparent frequency of note produced by the source is 400 cps. Then, its true frequency (in cps) is (velocity of sound in air = 360 m/s)

420

360

400

320

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Explanation

$n' = n (\frac{v + v_{O}}{v - v_{S}}) n \Rightarrow 400 = n (\frac{360 + 40}{360 - 40}) \Rightarrow n = 320 c p s$

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A siren emitting sound of frequency 500 Hz is going away from a static listener with a speed of 50 m/sec. The frequency of sound to be heard, directly from the siren, is :

434.2 Hz

589.3 Hz

481.2 Hz

286.5 Hz

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Explanation

$n^{'} = n (\frac{v}{v + v_{S}}) = 500 \times (\frac{330}{300 + 50}) = 434 .2 H z$

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A man sitting in a moving train hears the whistle of the engine. The frequency of the whistle is 600 Hz

The apparent frequency as heard by him is smaller than 600 Hz

The apparent frequency is larger than 600 Hz

The frequency as heard by him is 600 Hz

None of the above

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Explanation

Since there is no relative motion between the listener and source, hence actual frequency will be heard by listener.

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A whistle of frequency 500 Hz tied to the end of a string of length 1.2 m revolves at 400 rev/min. A listener standing some distance away in the plane of rotation of whistle hears frequencies in the range (speed of sound = 340 m/s)

436 to 586

426 to 574

426 to 584

436 to 674

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Explanation

The linear velocity of Whistle

$v_{S} = r ω = 1 .2 \times 2 π \frac{400}{60} = 50 m / s$

When Whistle approaches the listener, heard frequency will be maximum and when listener recedes away, heard frequency will be minimum

So, $n_{\max} = n (\frac{v}{v - v_{s}}) = 500 (\frac{340}{290}) = 586 H z$

$n_{\min} = n (\frac{v}{v + v_{s}}) = 500 (\frac{340}{390}) = 436 H z$

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A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f₁. If the train’s speed is reduced to 17 m/s, the frequency registered is f₂. If the speed of sound is 340 m/s then the ratio f₁/f₂ is

18/19

1/2

19/18

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Explanation

By using $n' = n (\frac{v}{v - v_{S}})$

⇒ $f_{1} = n (\frac{v}{v - v_{S}}) = n (\frac{340}{340 - 34}) = \frac{340}{306} n$

and $f_{2} = n (\frac{340}{340 - 17}) = n (\frac{340}{323})$

$\Rightarrow \frac{f_{1}}{f_{2}} = \frac{323}{306} = \frac{19}{18}$

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If source and observer both are relatively at rest and if speed of sound is increased then frequency heard by observer will :

Increases

Decreases

Can not be predicted

Will not change

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Explanation

No change in frequency.

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A source and an observer move away from each other with a velocity of 10 m/s with respect to ground. If the observer finds the frequency of sound coming from the source as 1950 Hz, then actual frequency of the source is (velocity of sound in air = 340 m/s)

1950 Hz

2068 Hz

2132 Hz

2486 Hz

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Explanation

$n^{'} = n (\frac{v - v_{O}}{v + v_{S}}) = n (\frac{340 - 10}{340 + 10}) = 1950$

$\Rightarrow n = 2068 H z$

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A whistle revolves in a circle with an angular speed of 20 rad/sec using a string of length 50 cm. If the frequency of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer, which is far away from the centre in the same plane ? (v = 340 m/s)

333 Hz

374 Hz

385 Hz

394 Hz

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Explanation

Minimum frequency will be heard, when whistle moves away from the listener.

$n_{\min} = n (\frac{v}{v + v_{s}})$ where $v = r ω = 0 .5 \times 20 = 1 m / s$

⇒ $n_{\min} = 385 (\frac{340}{340 + 10}) = 374 H z .$

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A Siren emitting sound of frequency 800 Hz is going away from a static listener with a speed of 30 m/s, frequency of the sound to be heard by the listener is (take velocity of sound as 330 m/s)

733.3 Hz

644.8 Hz

481.2 Hz

286.5 Hz

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Explanation

$n' = n (\frac{v}{v + v_{S}}) = 800 (\frac{330}{330 + 30}) = 733 .33 H z$ .

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A car sounding a horn of frequency 1000 Hz passes an observer. The ratio of frequencies of the horn noted by the observer before and after the passing of the car is 11 : 9. If the speed of sound is v, the speed of the car is

$\frac{1}{10} v$

$\frac{1}{2} v$

$\frac{1}{5} v$

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