Question

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.

Answer 1

Given `v^(2)=alpha-betay^(2)` …(i)
Differentiating it w.r.t. time t, we have,
`2v(dv)/(dt)=-beta2y(dy)/(dt)=-beta2(v)`
`:. (dv)/(dt)=-betay` …(ii)
It means, acceleration, `A=(dv)/(dt)=-betay`
As, `Apropy` and `-ve` sign shows that acceleration is directed towards mean position, so if the particle is left free, it will execute linear S.H.M.
Here, `omega^(2)=beta or omega=sqrt(beta)`
`:.` time period, `T=(2pi)/(omega)=(2pi)/(sqrt(beta))`
We know that, `v=0, ` when `y=r`.
From (i), `0=alpha-betar^(2) or r=sqrt((alpha)/(beta)), i.e., ` amplitude `=r=sqrt((alpha)/(beta))`