Question

Two sources A and B are sounding notes of
frequency 680Hz. A listener moves from A to
B with a constant velocity u. If the speed of
sound is 340 ms', what must be the value of
u so that he hears 10 beat/ sound?
a) 2 ms' b)
2.5 ms
3 ms' d) 3.5ms'
​

Answer 1

Answer:

option (a) is the correct answer

Explanation:

please mark as brainiest

Answer 2

Answer:

2.5m/s

Explanation:

Hi there... I'll give a brief explanation of doppler's effect first.

Since the person is moving from one source to the other with a constant speed, he would hear different frequencies from either of the sources. This phenomenon is called Doppler's effect.

now, in the given question, let frequency from A be $f_a$ and from B be $f_b$

We know,

$f_{apparent} =$ f[V±Vo] / [V±Vs]

Since he is moving away from A,

$f_{a}= f\frac{V-V_o}{V}$

(Vs=0)

similarly, he moves towards B

$f_b=f\frac{V+V_o}{V}$

also, we have been provided with beat frequency

Δf=10 bps          also, V=340m/s

fb-fa=10bps

solving, we get,

Vo=2.5m/s

I hope this helps you.

Please mark this the brainliest if it helped...