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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)) = -alphax………(i)`
We know `a = (d^(2)x)/(dt^(2)) = -omega^(2)x …….(ii)`
From Eq. `(i)` and `(ii)`, we have
`omega^(2) = alpha`
`omega = sqrt(alpha)`
or `(2pi)/(T) = sqrt(alpha) :. T = (2pi)/(sqrt(a))`
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