Question

Two sources A and B are sending notes of frequency 680 Hz. A listener moves from A and B with-a constant velocity u. If the speed of sound in air is 340 ms${}^{-1}$, what must be the value of u so that he hears 10 beats per second?

• A. 2.0 ms${}^{-1}$
• B. 2.5 ms${}^{-1}$
• C. 3.0 ms${}^{-1}$
• D. 3.5 ms${}^{-1}$

Answer 1

The correct option is B 2.5 ms${}^{-1}$

Listener go from A$\to$B with velocity (u) let the apparent frequency of sound from source A by listener
$n^{\prime }=n\frac{v-v_a}{v+v_s}$
Or $n^{\prime }=680\frac{340-u}{340+0}$
The apparent frequency of sound from source B by listener
$n"=n\frac{v+v_A}{v-v_S}=680\frac{340+u}{340-0}$
But listner he3ear 10 beats prer second.
Or $680\frac{340+u}{340}-680\frac{340-u}{340}=10$
Or $2340+u-340+u=10$
$u=2.5m{s}^{-1}$