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

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) \(\;\frac{{7A}}{3}\)

Detailed explanation

Let new amplitude is \(A'\)

initial velocity

\(v^{2}=\omega^{2}\left(A^{2}-\left(\frac{2 A}{3}\right)^{2}\right)\) \(...(1)\)

Where \(A\) is initial amplitude \(\&\, \omega\) is angular frequency.

Final velocity

\((3 v)^{2}=\omega^{2}\left(A^{\prime 2}-\left(\frac{2 A}{3}\right)^{2}\right)\) \(...(2)\)

From equation \(\&\) equation \(( 2)\)

\(\frac{1}{9}=\frac{A^{2}-\frac{4 A^{2}}{9}}{A^{2}-\frac{4 A^{2}}{9}}\)

\(A^{\prime}=\frac{7 A}{3}\)

Same subject

From JEE Mains