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A particle executes a SHM of angular vel...

A particle executes a SHM of angular velocity 2 rad/s and maximum acceleration of `8 m//s^(2)`. What is the path length of the oscillator ?

A

2 m

B

3 m

C

4 m

D

6 m

Text Solution

Verified by Experts

The correct Answer is:
C
`a = omega^(2)A therefore A= (a)/(omega^(2))=(8)/(4)=2`
`therefore` Path length=4 cm
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