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

Answer: (B)

$2.5 \, ms^{-1}$

Answer 2

Listener go from $A \to B$ with velocity $(u)$
let the apparent frequency of sound from source $A$ by listener
$n' = n \frac{v-v_{0}}{v+v_{s}}$
Or $n' = 680 \frac{340 -u}{340 + 0}$
The apparent frequency of sound from source $B$ by listener
$n" = n \frac{v+v_{0}}{v-v_{s} } = 680 \frac{340+u}{340-0}$
But listener hear $10$ beats per second.
Or $680 \frac{340 + u}{340} - 680 \frac{340 - u }{340} = 10$
Or $2340 + u - 340 + u = 10$
$u =2.5 \,ms^{-1}$