A point performs harmonic oscillations along a straight line with a period `T=0.60 s` and amplitude `a = 10.0 cm`. Find the mean velocity of the point averaged over the time interval during which it travels a distance `z//2`, starting from
`(a)` the extreme position.
`(b)` the equilibrium position.

Explanation

When a particle starts from an extreme position, it is useful to write the motion law as
`x=a cos omegat ….(1)`
(However `x` is the displacement from the equilibrium position )
It `t_(1)` be the time to cover the distane `a//2` then from `(1)`
`a-(a)/(2)=(a)/(2)=a cos omega t_(1)` or `cos omegat_(1)=(1)/(2) =cos ((pi)/(3))` (as `t_(1) lt T//4)`
Thus `t_(1)=(pi)/(3 omega)=(pi)/(3(2pi//T))=(T)/(6)`
As `x=a cos omega t`, so` v_(x)=-a omega sin omega t`
Thus `v=|v_(x)|=-v_(x)=a omega sin omegat`, for` tlet_(1)=T//6`
Hence sought mean velocity
`lt v gt = (int v dt)/(int dt)=int_(0)^(T//6)a(2pi//T)sin omegat dt //T//6=(3a)/(T)=(3a)/(T)=0.5 m//s`
`(b)` In this case, it is easier to write the motion law in the form `:`
`x=a sin omegat .......(2)`
If `t_(2)` be the time to cover the distance `a//2` , then from Eqn `(2)`
`a//2=a sin ((2pi)/(T))t_(2)` or `sin ((2pi)/(T))t_(2)=(1)/(2)=sin ((pi)/(6))`( as `t_(2)//4)`
Thus `(2pi)/(T)t_(2)=(pi)/(6)` or `t_(2)=(T)/(12)`
Differentiating Eqn `(2)` w.r.t. time, we get
`v_(x)=a omega cos omegat =a(2pi)/(T)cos ((2pi)/(T))t`
So, `v=|v_(lambda)|=a(2pi)/(T)cos((2pi)/(T))t`, for`t let_(2)=T//12`
Hence the sought mean velocity
`lt v gt =(int vdt)/(int dt)=(1)/((T//12))int_(0)^(T//12)a(2pi)/(T)cos ((2pi)/(T))t dt=(6a)/(T)=1m//s`