Question

A source of sound vibrating with a frequency 510 Hz is in between a stationary observer and a fort wall. If the source is moving towards wall with a velocity 3 m/s, the number of beats heard by the observer is (velocity of sound in air = 340 m/s)

• A. 3 beats/s
• B. 6 beats/s
• C. 9 beats/s
• D. 4.5 beats/s

Answer 1

Answer: (C)

9 beats/s

Apparent frequency, $f^{\prime }=\left(\frac{v}{v-v_s}\right)f$

$\Rightarrow f_1^{\prime }=\left(\frac{340}{340-3}\right)510$

Apparent frequency as seem by observer is:
$f_2^{\prime }=\left(\frac{340}{340+3}\right)510$

$\therefore f_1^{\prime }-f_2^{\prime }=\left(\frac{2\times 3\times 340}{(340{)}^2-(3{)}^2}\right)510$ $=9Hz$