Question

A particle performs S.H.M of amplitude A along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}$ A from mean position, its kinetic energy gets increased by an amount $\frac{1}{2}m{\omega }^2{A}^2$ due to an impulsive force. Then its new amplitude becomes.

• A. $\frac{\sqrt{5}}{2}$A
• B. $\frac{\sqrt{3}}{2}$A
• C. $\sqrt{2}$A
• D. $\sqrt{5}$A

Answer 1

The correct option is C $\sqrt{2}$A

If the particle performs SHM with amplitude A along the straight line, than any point the total energy should be

$E=K+P=\frac{1}{2}K{A}^2=\frac{1}{2}m{w}^2{A}^2$

If the additional kinetic energy of magnitude is considered,

$E=E+\frac{1}{2}m{w}^2{A}^2=\frac{1}{2}m{w}^2{A}^2+\frac{1}{2}m{w}^2{A}^2$

$⟹\frac{1}{2}m{w}^2{A}^2=\frac{1}{2}m{w}^2\left(2A^2\right)⟹{A^{\prime }}^2=2A^2$

$⟹A^{{}^{\prime }}=\sqrt{2A}$