wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A detector is released from rest over a source of sound of frequency f0=103 Hz. The frequency observed by the detector at time t is plotted in the graph. The speed of sound in air is (g=10 m/s2)

A
330 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
350 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
300 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
310 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 300 m/s
Initially the detector (observer) is at rest.
Velocity of observer after 30 sec,
v0=0+gt=10×30=300 m/sec
Given : f0=1000 Hz and f=2000 Hz using doppler's effect in case an observer moves towards the stationary source;
f=f0[v+v0v]
(where v is the speed of sound in air)
2000=1000[v+300v]
v=300 m/sec

flag
Suggest Corrections
thumbs-up
8
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Superposition Recap
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon