Question

Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0m s−1 so that he approaches one tuning fork and recedes from the other figure. Find the beat frequency observed by the listener. Speed of sound in air = 332 m s−1.

Answer 1

Beat Frequency is 4.6 Hz

Explanation:

Given:

Speed of sound in air v = $332\ ms^-^1$

The velocity of the observer $v_0 = 3\ ms^-^1$

The velocity of the source $v_s$ = 0

Frequency of the tuning forks $f_0$ = 256 Hz

• Apparent Frequency (f_1)that the man heard while running towards the tuning forks -

$f_1 = (\frac {v +v_0}{v}) \times f_0$

• Lets Substitute the values in equation - we get:

$f_1 = (\frac {332 + 3}{332}) \times 256$  

• When man is running away from forks- Apparant frequency becomes  

$f_2 = (\frac {v - v_0}{v}) \times 256$

= 253.7 Hz

• So, Beats produced by them  

= $f_2 - f_1$

= 258.3 - 253.7 = 4.6 Hz

Answer 2

Answer:

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