There is a source of sound (a tuning fork) moving towards a reflecting wall will a speed of 30 m s^(-1). The velocity of the sound in air is 330 m…

Question

There is a source of sound (a tuning fork) moving towards a reflecting wall will a speed of

30 m s^{- 1}

. The velocity of the sound in air is

330 m s^{- 1}

and the frequency of the tuning fork is

600 Hz

. The observer is between the source and the wall and is moving with some velocity

V_{1}

towards the wall. The beat frequency heard by the observer is

4 Hz

. If the tuning fork is waxed, the frequency beats heard by the observer becomes

3 Hz

. If the new frequency of tuning fork is

k

then

\frac{k}{50}

is

MARKS Content Team · Accepted Answer

Correct answer: 9. Apparent frequency of source =n_0(c-v_1)/(c-v_2)=n^(′) Apparent frequency of reflected sound =n_0(c+v_1)/(C-V)=n^(′′) Beat frequency =n^(′′)-n^(′)=2n_0V_1/(C-V_2)=(2×600×V_1)/(330-30)=4 Hz ∴ V_1= 1 m s^(-1) When the tuning fork is waxed let its frequency become k Beat frequency =2kv_1/(C-V_2)=(2×k×1)/(330×30) =3 Hz ∴ k=450 Hz= new frequency of tuning fork. k50=9

Solution

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