Question

A particle performs S.H.M. with amplitude \( A \). Its speed is tripled at the instant when it is at a distance of \( \frac{2A}{3} \) from the mean position. The new amplitude of the motion is

• A. \( \frac{5A}{3} \)
• B. \( \frac{7A}{3} \)
• C. \( \frac{2A}{3} \)
• D. \( \frac{A}{3} \)

Hint:

In S.H.M., a change in speed without changing position indicates a change in total energy, which directly affects the amplitude.

Answer 1

The Correct Option is B

Step 1: Write the expression for speed in S.H.M.
The speed of a particle performing S.H.M. at displacement \( x \) is given by: \[ v = \omega \sqrt{A^2 - x^2} \]

Step 2: Apply the given condition.
At displacement \( x = \frac{2A}{3} \), the initial speed is: \[ v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\omega A \sqrt{5}}{3} \]

Step 3: Use the condition of tripled speed.
When the speed is tripled, the new speed becomes: \[ v' = 3v = \omega A \sqrt{5} \] Let the new amplitude be \( A' \). Then, \[ v' = \omega \sqrt{A'^2 - \left(\frac{2A}{3}\right)^2} \] \[ \omega A \sqrt{5} = \omega \sqrt{A'^2 - \frac{4A^2}{9}} \]

Step 4: Solve for the new amplitude.
\[ 5A^2 = A'^2 - \frac{4A^2}{9} \] \[ A'^2 = \frac{49A^2}{9} \Rightarrow A' = \frac{7A}{3} \]

Step 5: Final conclusion.
The new amplitude of the motion is \( \frac{7A}{3} \).