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The equation of motion of a particle of ...

The equation of motion of a particle of mass `1g` is `(d^(2)x)/(dt^(2)) + pi^(2)x = 0`, where `x` is displacement (in m) from mean position. The frequency of oscillation is (in Hz)

A

`(1)/(2)`

B

2

C

`(sqrt(2)g)/(2pi)`

D

`(1)/(5sqrt(10))`

Text Solution

Verified by Experts

The correct Answer is:
A
`(d^(2)x)/(dt^(2))+pi^(2)x=0`
`rArr ` Compare with `(d^(2)x)/(dt^(2))+omega^(2)x=0`
so `omega = pi`
So `f=(omega)/(2pi)=(pi)/(2pi)=(1)/(2)Hz` `[Soln. `made by SSI Sir`]`
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