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A particle excutes simple harmonic motio...

A particle excutes simple harmonic motion with an amplitude of 5 cm. 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

`7/3pi`

B

`8pi/3`

C

`3/8pi`

D

`4pi/3`

Text Solution

Verified by Experts

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