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

The correct option is D $\frac{7A}{3}$

$\begin{array}{c}m{w}^2=k\\ Totalinitialenergy=\frac{1}{2}k{A}^2\\ atx=\frac{2A}{3},potentialenergy=\frac{1}{2}k{\left(\frac{2A}{3}\right)}^2\\ =\left(\frac{1}{2}k{A}^2\right)\left(\frac{4}{9}\right)\\ Kineticenergyat\left(x=\frac{2A}{3}\right)is=\left(\frac{1}{2}k{A}^2\right)\left(\frac{5}{9}\right)\\ Ifspeedistriled,newkineticenergy=\frac{1}{2}k{A}^2\cdot \frac{5}{9}=\frac{5}{2}k{A}^2\\ \therefore Newtotalenergy=\frac{5}{2}k{A}^2+\frac{1}{2}k{A}^2\left(\frac{4}{9}\right)\\ ifnextamplitude=A^{\prime }\\ then\frac{1}{2}kA{}^{\prime 2}\left(\frac{49}{9}\right)\\ \Rightarrow A^{\prime }=\frac{7A}{3}\end{array}$

Hence,

option $\left(D\right)$ is correct answer.