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: (D)

$\frac{7A}{3}$

Answer 2

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