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A particle is executing a simple harmoni...

A particle is executing a simple harmonic motion its maximum acceleration is a and maximum velocity is `beta` .Then its time of vibration will be

A

`(2pialpha)/(beta)`

B

`(2pi beta)/(alpha)`

C

`2pialphabeta`

D

`(pibeta)/(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
A

`alpha=A omega`
`beta=omega^(2) A `
`T=-(2pi alpha)/(beta)`
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RESONANCE-SIMPLE HARMONIC MOTION-Exercise
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  16. The potential energy of a particle execuring S.H.M. is 2.5 J, when its...

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  17. A body of mass m is suspended from three springs as shown in figure. I...

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  18. One mass m is suspended from a spring. Time period of oscilation is T....

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