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

$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 $

Answer 2

As, the velocity of a particle executing SHM at any instant, is the time rate of change of its displacement at that instant.

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

Where, \(\omega\) is the angular frequency, A is the amplitude, x is the displacement of the particle.

Let us assume that the new amplitude of the motion is ‘\(A\)’.

So, if A is the initial amplitude and ω is the angular frequency. Then the initial velocity of a particle performs SHM,

⇒ \(v^{2}=\omega ^{2}[{A^{2}-(\frac{2A}{3})^{2}}\)---(i)

So, now the final velocity of the particle is given by,

⇒ \((3v)^{2}=\omega ^{2}[{A'^{2}-(\frac{2A}{3})^{2}}\)---(ii)

 So now, the ratio of equation (i) and (ii) can be written as,

⇒ \(\frac{v^{2}}{(3v)^{2}}=\frac{{A^{2}-\frac{4A}{9}}}{{A'^{2}-\frac{4A}{9}}}\)

⇒ \(\frac{v^{2}}{9}=\frac{\frac{9A^{2}-4A^{2}}{9}}{{{A'^{2}-\frac{4A}{9}}}}\)

⇒ \(\frac{1}{9}=\frac{\frac{5A^{2}}{9}}{{{A'^{2}-\frac{4A}{9}}}}\)

⇒ \(A'^{2}=\frac{45A^{2}+4A^{2}}{9}\)

⇒ \(A'^{2}=\frac{49A^{2}}{9}\)

⇒ \(A'^{2}=\frac{7A}{3}\)

Therefore, the new amplitude of the motion is \(\frac{7A}{3}\).

Hence the correct option is ‘D’.