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A particle executes simple harmonic moti...

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

A

`sqrt(2)pi`

B

`16sqrt(2)pi`

C

`24sqrt(2)pi`

D

`32sqrt(2)//pi`

Text Solution

Verified by Experts

The correct Answer is:
D
`T=16` sec
At t=2sec
Displacement, `y=Asinomegat`
At t=2sec
`y.=Asin((2pi)/(T)) t=Asin((2pi)/(16)xx2)=(A)/(sqrt2)`
Velocity `(v)=omega sqrt(A^(2)-y^(2))`
`implies 4=(2pi)/(16)sqrt(A^(2)-((A)/(sqrt2))^(2))implies A=(32sqrt(2))/(pi)` m.
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