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A particle executes simple harmonic moti...

A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time in seconds is :

A

`(7)/(3)pi`

B

`(8pi)/(3)`

C

`(3)/(8)pi`

D

`(4pi)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
`|omega^(2)x|=|omega sqrt(A^(2)-x^(2))|`
`rArr omega^(2)x^(2)=A^(2)-x^(2) rArr omega=(3)/(4)`
`T=(8pi)/(3)`
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