1.

A.S.H.M. oscillator has period of 0.1 s and amplitude of 0.2 m. The maximum velocity is given by :

Answer»

`100" m s"^(-1)`
`4PI" m s"^(-1)`
`100pi" m s"^(-1)`
`20PI" m s"^(-1)`

Solution :`omega=(2pi)/(T)`
and `v_("max")=r omega`.
`v_("max")=0.2xx(2pi)/(0.1)=4pi" m"//"s"`
So the correct choice is (B).


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