Question

Two sound sources are moving in opposite directions with velocities $v_1$ and $v_2(v_1>v_2)$. Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of $v_1–v_2$ so that the beat frequency observed by the observer is 6 Hz? Speed of sound v = 300 m/s. Given that $v_1$ and $v_2$ ≪ v.

• A. 1 m/s
• B. 2 m/s
• C. 3 m/s
• D. 4 m/s

Answer 1

The correct option is B 2 m/s

$f_1=900\left(\frac{300}{300+v_1}\right)\cong 900{(1+\frac{r_1}{300})}^{-1}900-3v_1$
Similarly,
$f_2=900\left(\frac{300}{300+v_2}\right)=900-3v_2{f}_2-f_1=6\therefore 3(v_1-v_2)=6\therefore 3(v_1-v_2)=6or{v}_1-v_2=2m/s$