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For a particle performing linear SHM, it...

For a particle performing linear SHM, its average speed over one oscillation is (where, a= amplitude of SHM, n=frequency of oscillation)

A

`(A omega)/(pi)`

B

`(2 A omega)/(pi)`

C

Zero

D

`(2pi)/(A omega)`

Text Solution

Verified by Experts

The correct Answer is:
B
Average speed `=("Dis"tan"ce travelled")/("Time")=(4A)/(T)`
but `T=(2pi)/(omega) " " therefore A.S.=(4A xx omega)/(2pi)=(2Aomega)/(pi)`
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