Question

Two sources $A$ and $B$ are sounding notes of frequency $660Hz$ A listener moves from $A$ to $B$ with a constant velocity $u.$ If the speed of sound is $330m/s$ what must be the value of $u$ so that he hears $8$ beats per second?

• A. $2.8m/s$
• B. $2m/s$
• C. $3.0m/s$
• D. $3.5m/s$

Answer 1

The correct option is B $2m/s$

Apparent frequency of sound heard by the observer from source is

$n_1=\left(\frac{v-v_o}{v-v_s}\right)n$

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

Apparent frequent of sound heard by the observe from source

$n_2=\left(\frac{v+v_o}{v-v_o}\right)n$

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

No. of beats$=8$

$n_2-n_1=8$

$\left(\frac{v+u}{v-v_s}\right)n-\left(\frac{v-u}{v+v_s}\right)n=8$ $v=300,v_s=0,n=660$

$\Rightarrow \left(\frac{330+u}{300-0}\right)\left(600\right)-\left(\frac{300-u}{330+0}\right)660=8$

$\therefore \frac{2\times 6604}{330}=8$

$4u=8$

$u=2$.