Question

A particle performs simple harmonic motion with amplitude \( A \). Its speed is tripled at the instant that it is at a distance \( \frac{2A}{3} \) from the equilibrium position. The new amplitude of the motion is:

• A. \( A\sqrt{3} \)
• B. \( \frac{7A}{3} \)
• C. \( \frac{A}{3}\sqrt{41} \)
• D. \( 3A \)

Hint:

In SHM, when velocity changes, use energy conservation or velocity equations to relate the change in speed to the amplitude. Here, tripling the speed leads to a proportional change in amplitude.

Answer 1

The Correct Option is B

In this problem, we are tasked with determining the new amplitude \( A' \) when the velocity of a particle performing simple harmonic motion triples. The velocity \( V \) of a particle in simple harmonic motion is given by the equation: \[ V = \omega \sqrt{A^2 - x^2}, \] where: \[ \omega = \text{angular frequency}, \quad A = \text{amplitude}, \quad x = \text{displacement from equilibrium}. \] Step 1: Initial velocity at displacement \( x = \frac{2A}{3} \) The initial velocity is given by: \[ V = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2}. \] Simplify: \[ V = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3}. \] Step 2: Velocity when speed is tripled The final velocity is three times the initial velocity, so: \[ 3V = \omega \sqrt{A'^2 - \left(\frac{2A}{3}\right)^2}. \] Substitute the expression for \( V \): \[ 3 \left(\omega \frac{\sqrt{5}A}{3}\right) = \omega \sqrt{A'^2 - \frac{4A^2}{9}}. \] Simplify: \[ \omega \sqrt{5}A = \omega \sqrt{A'^2 - \frac{4A^2}{9}}. \] Step 3: Solve for \( A' \) Square both sides to eliminate the square roots: \[ 5A^2 = A'^2 - \frac{4A^2}{9}. \] Rearrange the equation: \[ A'^2 = 5A^2 + \frac{4A^2}{9}. \] Simplify: \[ A'^2 = \frac{45A^2}{9} + \frac{4A^2}{9} = \frac{49A^2}{9}. \] Now, take the square root of both sides: \[ A' = \frac{7A}{3}. \] Final Answer: \[ \boxed{\frac{7A}{3}}. \]