Question

A particle moving with SHM in a straight line has a speed of `6m//s` with when 4m from the centre of oscillation and a speed of `8m//s` when 3m from the centre of oscillation . Find the amplitude of oscillation and the shortest time taken by the particle in moving from the extremen position to a point mid way between the extreme position and the centre.

Answer 1

`V^(2)=omega^(2)(r^(2)-y^(2))`
(i) `6^(2)=omega^(2)(r^(2)--4^(2))` (ii) `omega^(2)(r^(2)-3^(2))`
or `(64)/(36)=(r^(2)-9)/(r^(2)-16)`
On solving, `r=+-5m and omega =2s^(-1)`
For the given displacement, `x=(r)/(2),t=?`
Asx `x=r cos omega t`
`:. (r)/(2)=r cos 2t or cos 2t=(1)/(2)=cos((pi)/(3))`
or `2t=(pi)/(3)or t=(pi)/(6)s`