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For a particle in S.H.M. if the amplitud...

For a particle in `S.H.M.` if the amplitude of displacement is `a` and the amplitude of velocity is `v` the amplitude of acceleration is

A

`va`

B

`(v^(2))/(a)`

C

`(v^(2))/(2a)`

D

`(v)/(a)`

Text Solution

Verified by Experts

The correct Answer is:
B
`v_("max") = a omega` and
Maximum acceleration = `omega^(2) a`
`= ((v)/(a))^(2) a = (v^(2))/(a)`
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