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If the maximum speed of a SHO is pi ms^(...

If the maximum speed of a SHO is `pi ms^(-1)`. Its average speed during one oscillations is

A

`(pi)/(2)ms^(-1)`

B

`(pi)/(4)ms^(-1)`

C

`pims^(-1)`

D

`2ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`V_(m)=Aomega=pi,pi=AomegaimpliesA=(pi)/(omega)`
Average speed `=(4A)/(T)=(4)/(T)xx(pi)/(omega)=(4omega)/(2pi)xx(pi)/(omega)`
`=2ms^(-1)`
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