A particle moving in a straight line has velovity v given by `v^(2)=alpha-betay^(2)` , where `alpha` and `beta` are constants and y is its distance from a ficed point in the line. Show that the motion of particle is SHM. Find its itme period and amplitude.

Explanation

Given , `v^(2)=alpha-betay^(2)` …(i)
Differentiating it w.r.t. time t, we have
`2v(dv)/(dt)=-beta2y(dv)/(dt)=-betay`
`:.` Acceleration, `A=(dv)/(dt)=-betay` …(ii)
As `Apropy` and `-ve` sign shows that acceleration is directed towards the mean position, so if the particle is left free, is will execute SHM.
Comparing (ii), with `A=-omega^(2)y`, we have
`omega^(2)=beta or omega=sqrt(beta)`
Time period, `T=(2pi)/(omega)=(2pi)/(sqrt(beta))`
We known that `v=0, ` when `y=r` From (i),
`o=alpha-beta r^(2)or r=sqrt(alpha//beta)`
`:.` Amplitude `=r=sqrt(alpha//beta)`