Question

A particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from equilibrium position. The new amplitude of the motion is :

• A. $\frac{A}{3}\sqrt{41}$
• B. 3A
• C. $A\sqrt{3}$
• D. $\frac{7A}{3}$

Answer 1

Answer: (C)

$m{\omega }^2=k$
Total initial energy $=\frac{1}{2}k{A}^2$
at $x=\frac{2A}{3},$ potential energy $=\frac{1}{2}k{\left(\frac{2A}{3}\right)}^2=\left(\frac{1}{2}k{A}^2\right)\left(\frac{4}{9}\right)$
Kinetic energy at $\left(x=\frac{2A}{3}\right)=\left(\frac{1}{2}k{A}^2\right).\left(\frac{5}{9}\right)$
If speed is tripled, new Kinetic energy $=\frac{1}{2}k{A}^2.\frac{5}{9}=\frac{5}{2}k{A}^2$
$\therefore$ New total energy $=\frac{5}{2}k{A}^2+\frac{1}{2}k{A}^2\left(\frac{4}{9}\right)=\frac{k{A}^2}{2}\left(\frac{49}{9}\right)$
If next amplitude = A' ; then $\frac{1}{2}k{A}^{\prime 2}=\frac{1}{2}k{A}^2\left(\frac{49}{9}\right)\Rightarrow A^{\prime }=\frac{7}{3}A$