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The displacement-time relation for a par...

The displacement-time relation for a particle can be expressed as `x=0.5[cos^(2)(npit)-sin^(2)(npit)]`. This relation shows that

A

the particle execute SHM with amplitude 0.5 m

B

the particle execute SHM with a frequency n times that of a second pendulum

C

the particle executes SHM and the velocity in its mean position is `(3.142n)m*s^(-1)`

D

the particle does not execute SHM at all.

Text Solution

Verified by Experts

The correct Answer is:
A, C
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