1.

A particle is moving in a straight line with S.H.M. of amplitude r. At a distance s from the mean position of motion, the particle receives a blow in the direction of motion which instantaneously doubles the velocity. Find the new amplitude.

Answer»

Velocity v = ω\(\sqrt{r^2\,-\,y^2}\)

At y = s, let v = v0 then,

\(v^2_0\) = ω2(r2 − s2 ) …(i)

Due to blow, the new velocity at y = s is

V = 2v0, r = r′

So (2v0)2 = ω2(r'2 − s2 ) …(ii)

Dividing (ii) by (i),

 \(4=\frac{r'^2-s^2}{r^2-s^2}\)

On solving r′= \(\sqrt{4r^2\,-\,3s^2}\)



Discussion

No Comment Found

Related InterviewSolutions