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For a particle performing `SHM`, equation of motion is given as `(d^(2))/(dt^(2)) + 4x = 0`. Find the time period

A

`2pi`

B

`(1)/(3)pi`

C

`(2)/(3)pi`

D

`4 pi`

Text Solution

Verified by Experts

The correct Answer is:
C

Given, `(d^(2)x)/(dt^(2))=-9x`
Comparing with `(d^(2)x)/(dt^(2))=-omega^(2)x rArr = 9, omega = 3`
`therefore` Time period `= (2pi)/(omega)=(2pi)/(3)=(2)/(3)pi`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-OSCILLATIONS-EXERCISE 2
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