Question

A source of sound of frequency $256Hz$ is moving rapidly towards a wall with a velocity of $5m/s$. The speed of sound is $330m/s$. If the observer is between the wall and the source, then beats per second heard will be

• A. Zero
• B. $3.9Hz$
• C. $6.7Hz$
• D. $7.8Hz$

Answer 1

The correct option is D $7.8Hz$

The observer will hear two sounds, one directly from source and other from reflected image of sound.

Sound heard directly from source,
$n_1=n\left(\frac{v}{v-v_s}\right)$

Sound heard by reflected image of sound,
$n_2=n\left(\frac{v}{v+v_s}\right)$

Here,
$v=$ speed of sound $=330m/s$
$n=$ frequency of source $=256Hz$
$v_s=$ velocity of source $=5m/s$

Number of beats heard per sec
$=n_1-n_2$

$=n\left(\frac{v}{v-v_s}\right)-n\left(\frac{v}{v+v_s}\right)$

$=\frac{2nv{v}_s}{v^2-v_s^2}$

$=\frac{2nv{v}_s}{v^2-v_s^2}$

Substitute the values,

$=\frac{2\times 256\times 330\times 5}{(330{)}^2-(5{)}^2}$

$=\frac{2\times 256\times 330\times 5}{(330+5)\times (330-5)}$

[ Note: $a^2-b^2=(a+b)(a-b)$ ]

$=\frac{2\times 256Hz\times 330\times 5}{335\times 325}$

$=7.8Hz$

Final answer: (a)