1.

If the maximum speed of a SHO is `pi ms^(-1)`. Its average speed during one oscillations isA. `(pi)/(2)ms^(-1)`B. `(pi)/(4)ms^(-1)`C. `pims^(-1)`D. `2ms^(-1)`

Answer» Correct Answer - D
`V_(m)=Aomega=pi,pi=AomegaimpliesA=(pi)/(omega)`
Average speed `=(4A)/(T)=(4)/(T)xx(pi)/(omega)=(4omega)/(2pi)xx(pi)/(omega)`
`=2ms^(-1)`


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