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If the differential equation for a simpl...

If the differential equation for a simple harmonic motion is `(d ^(2) y )/(dt ^(2)) + 2 y =0,` the time period of the motion is

A

`pi sqrt2 s`

B

`(sqrt2)/(pi)`

C

`(pi)/(sqrt2) s`

D

`2 pi s `

Text Solution

Verified by Experts

The correct Answer is:
A
Given,
`(d ^(2) Y)/(dt ^(2)) + 2y =0 " "...(i)`
But, we know that
`(d ^(2) y)/( dt ^(2)) + omega ^(2) y =0" "...(ii)`
Comparing both equation, we get
`omega ^(2) =2`
`implies omega = sqrt2`
The periodic time
` T = (2pi)/(omega )= ( 2pi )/(sqrt2)`
`= sqrt2 pi = pi sqrt2 s`
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