Questions
A source and an observer approach each other with same velocity 50 m/s. If the apparent frequency is 435 sec–1, then the real frequency is
a
320 sec–1
b
360 sec–1
c
390 sec–1
d
420 sec–1
detailed solution
Correct option is A
n'=nv+vOv−vS; Here v = 332 m/s and v0=vs=50 m/s⇒435=n332+50332−50⇒n=321.12 sec−1≈320 sec-1Talk to our academic expert!
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