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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

`(3)/(8) pi`

B

`(4 pi)/(3)`

C

`(7)/(3) pi`

D

`(8 pi)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
` v = omega sqrt(5^(2) - 4^(2)) = 3 omega`
` a = omega ^(2) xx 4 `
`|a| = |v| `
` 4 omega ^(2) = 3 omega rArr omega = (3)/(4) = (2 pi)/(T) rArr T = ( 8 pi)/(3) ` sec
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