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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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