Question

A particle performs S.H.M. of amplitude $A$ along a straight line. When it is t 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. $\sqrt{5}A$
• B. $\frac{\sqrt{5}}{2}A$
• C. $\frac{\sqrt{3}}{2}A$
• D. $\sqrt{2}A$

Answer 1

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

Given,
Amplitude $=A$
The distance from the mean position,
$x_1=\frac{\sqrt{3}}{2}A$
Increase in the kinetic energy
$\Delta K=\frac{1}{2}m{\omega }^2{A}^2$
Let the new amplitude be $A$
Whenever the particle performs simple Harmonic Motion with amplitude $A$ along the straight line then at any point the total energy will be,
$E=K+U\Rightarrow E=\frac{1}{2}K{A}^2\Rightarrow E=\frac{1}{2}m{\omega }^2{A}^2$
If additional Kinetic energy is taken into consideration, new energy will be,
$\Rightarrow E^{\prime }=E+\frac{1}{2}m{\omega }^2{A}^2\Rightarrow E^{\prime }=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}m{\omega }^2{A}^2\Rightarrow \frac{1}{2}m{\omega }^2{A}^{\prime 2}=\frac{1}{2}m{\omega }^{2}2A^2\Rightarrow A^{\prime 2}=2A^2\Rightarrow A=\sqrt{2}A$