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The velocity of a particle executing a s...

The velocity of a particle executing a simple harmonic motion is `13m*s^(-1)`, when its distance from the equilibrium position (Q) is 3 m and its velocity is `12m*s^(-1)`, when it is 5 m away from Q. The frequency of the simple harmonic motion is

A

`(5pi)/8`

B

`5/(8pi)`

C

`(8pi)/5`

D

`(8)/(5pi)`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that, `v=omegasqrt(A^(2)-x^(2))`
`13=omegasqrt(A^(2)-3^(2))" "...(1)`
and `12=omegasqrt(A^(2)-5^(2))" "...(2)`
Solving (1) and (2), we get, `omega=5/(8pi)`
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