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A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre is

A

`sqrt(2)pi`

B

`16sqrt(2)pi`

C

`32sqrt(2)//pi`

D

`4//pi`

Text Solution

Verified by Experts

The correct Answer is:
C
`y_(1) = A sin ((2pi)/(T)xxt), V = omega sqrt(A^(2) - y_(1)^(2))`
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