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If a simple harmonic motion is represent...

If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2))+alphax=0`, its time period is

A

`2pi/alpha`

B

`2pi/sqrt(alpha)`

C

`2pialpha`

D

`2pisqrt(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(d^(2)x)/(dt^(2))+ax=0`
Comparing it with the standard equation of SHM, we get : `(d^(2)X)/(dt^(2))+omega^(2)X=0`
We have `omega^(2)=alpha`
Time period `T=(2pi)/(omega)=(2pi)/(sqrtalpha)`.
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