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What is the maximum acceleration of the ...

What is the maximum acceleration of the particle doing the SHM `gamma=2sin[(pit)/2+phi]` where gamma is in cm?

A

`(pi)/(2) cms^(-2)`

B

`(pi^(2))/(2) cms^(-2)`

C

`(pi)/(4) cms^(-2)`

D

`(pi^(2))/(4) cms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`omega=(pi)/(2) "rad" s^(-1)` and A=2 cm
`therefore` Maximum acceleration `=omega^(2)A=(pi^(2))/(2) cms^(-2)`
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