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Maximum speed of a particle in simple ha...

Maximum speed of a particle in simple haromonic motion is `v_("max")`. Then average speed of a particle in one complete oscillation is equal to

A

`(v_("max"))/2`

B

`(v_("max"))/(pi)`

C

` (piv_("max"))/2`

D

`(2v_("max"))/(pi)`

Text Solution

Verified by Experts

The correct Answer is:
D
`V_("max")=omegaA=(2pi)/T A`
`ltV_("avg")gt=(4A)/T=4((V_("max"))/(2pi))=(2V_("max"))/(pi)`
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