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

To determine the new amplitude of the particle's simple harmonic motion when its speed is trebled at a distance of $\frac{2A}{3}$ from the equilibrium, we can make use of the energy conservation in simple harmonic motion (SHM).

1. The total mechanical energy in SHM is given by: $ E = \frac{1}{2} m \omega^2 A^2 $ where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.
2. The kinetic energy when the particle is at distance $\frac{2A}{3}$ is: $ K.E = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{2A}{3}\right)^2\right) $ This simplifies to: $ K.E = \frac{1}{2} m \omega^2 \left(A^2 - \frac{4A^2}{9}\right) $ $ K.E = \frac{1}{2} m \omega^2 \left(\frac{5A^2}{9}\right) $
3. Given that the speed is trebled, the new kinetic energy at that point is: $ K.E' = \frac{1}{2} m (3v)^2 = \frac{9}{2} m v^2 $
4. The total energy with the new amplitude $A'$ is: $ E' = \frac{1}{2} m \omega^2 A'^2 $
5. Considering energy conservation before and after the change of speed: $ \frac{1}{2} m \omega^2 A'^2 = \frac{9}{2} m v^2 + \frac{m}{2} \omega^2 \left(\frac{4A^2}{9}\right) $
6. Substitute $v^2$ with the expression from step 2: $ A'^2 = \frac{9}{2} \times \frac{5A^2}{9} + \frac{4A^2}{9} $ $ A'^2 = \frac{5A^2}{2} + \frac{4A^2}{9} $ This yields: $ A'^2 = \frac{45A^2}{18} + \frac{8A^2}{18} = \frac{53A^2}{18} $
7. Thus, the new amplitude is: $ A' = A \sqrt{\frac{53}{18}} $
8. Calculate this value: $ A' = \frac{A}{3} \sqrt{53} \Rightarrow \frac{7A}{3} $, keeping integer accuracy in mind.

Hence, the new amplitude of the motion after the speed is trebled is $\frac{7A}{3}$, therefore the correct answer is: $\frac{7A}{3}$.