Discussion Forum

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that  it is at  a distance 

$\frac{2}{3}$  A   from the equilibrium position. The new amplitude  of the  motion is 

Answer:

Explanation:

$v= \omega \sqrt{A^{2}-x^{2}}$

At  $x=\frac{2A}{3}$

$v=\omega\sqrt{A^{2}-(\frac{2A^{}}{3})^{2}}=\frac{\sqrt{5}}{3}\omega A$

As, velocity is trebled , hence v'=$\sqrt{5}A\omega$

This leads to new amplitude A'

$\omega\sqrt{A'^{2}-(\frac{2A}{3})^{2}}=\sqrt{5}A\omega$

$\Rightarrow \omega^{2}[A'^{2}-\frac{4A^{2}}{9}]=5A^{2}\omega^{2}$

$\Rightarrow A'^{2}= 5A^{2}+\frac{4}{9}A^{2}=\frac{49}{9}A^{2}$

$A'=\frac{7}{3}A$