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A particle is executing SHM with an ampl...

A particle is executing SHM with an amplitude of 2 m. The difference in the magnitudes of its maximum acceleration and maximum velocity is 4. The time period of its oscillation is

A

2s

B

`(7)/(22)s`

C

`(22)/(7)s`

D

`(4)/(7)s`

Text Solution

Verified by Experts

The correct Answer is:
C
`A omega^(2) - A omega = 4`
`2 omega^(2) - 2 omega = 4 rArr omega^(2) - omega - 2 = 0`
`rArr omega = 2`
`T = (2pi)/(2) = pi = (22)/(7)s`
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