A tiny mass performs S.H.M along a straight line with a time period of T=0.60sec and amplitude A=10.0cm. Calculate the mean velocity (in m/sec) in the time to displace by `A/2`.

Answer

1

Explanation

Time taken to travel from y=0 to `y=(A)/(2)` is found using `Y=Asinomegat`
`(A)/(2)=Asin((2pi)/(T)xxt)`
`(2pi)/(T)xxt=(pi)/(6)`
`Rightarrow t=(T)/(12)sec`
`v_(av)=(int_(0)^(T//12)vdt)/(int_(0)^(T//12)dt)=(int_(0)^(T//12)omegaAcosomegadt)/(((T)/(12)))`
`(12)/(T)omegaA|(sinomegat)/(omega)|_(0)^(T//12)=(12)/(T)A["sin"(2pi)/(T)xx(T)/(12)-sin0]`
` v_(ab)=(6xxA)/(T)=(60)/(0.60)cm//sec`
`Rightarrow v_(ab)=100cm//sec=1m//sec`