Question

Two sound sources of frequency 360 Hz each,
one moving toward observer, while second moving
away from observer with same speed 5.5 m/s, then
number of beats produced per second is (v = 330
m/s)
(1) 2
(2) 18
(3) 12
(4) 25​

Answer 1

answer : option (3) 12

explanation : from Doppler's effect,

frequency observed due to first source, f1 = f[v/(v - x)]

where, frequency of the source, f = 360 Hz, speed of sound, v = 330 m/s, speed of source, x = 5.5 m/s

so, f1 = 360 [330/(330-5.5)] = 360 × 330/324.5 Hz = 366 Hz

frequency observed due to 2nd source, f2 = f[v/(v + x)]

= 360 × 330/(330 + 5.5)

= 354 Hz

now number of beats per second = change in frequencies

= |366 - 354|

= 12

hence, number of beats per second is 12

Answer 2

Answer:12

Explanation: