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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′= \(\sqrt{4r^2\,-\,3s^2}\) |
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