JEE Main · 2016 · 03 Apr 2016 · Offline · Elasticity & SHM

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance

\frac{2 A}{3}

from equilibrium position. The new amplitude of the motion is :

Answer: (C)

Solution

In SHM the speed at displacement

x

v = ω A^{2} - x^{2}

. At

x = \frac{2 A}{3}

v = ω A^{2} - \frac{4 A ^{2}}{9} = \frac{ω A 5}{3}

. Trebling it to

v_{1} = 3 v = ω A 5

must be matched by a new amplitude

A_{1}

at the same point:

ω A 5 = ω A_{1}^{2} - \frac{4 A ^{2}}{9}

. Squaring,

5 A^{2} = A_{1}^{2} - \frac{4 A ^{2}}{9}

, so

A_{1}^{2} = \frac{49 A ^{2}}{9}

and

A_{1} = \frac{7 A}{3}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option