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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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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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