Question

Calculate the ratio of potential energy to kinetic energy at time \(t=\dfrac{T}{6}\) for a particle starting SHM from its mean position.

• A. \(1:1\)
• B. \(3:1\)
• C. \(1:3\)
• D. \(3:2\)

Hint:

In SHM starting from mean position, displacement is \(x=A\sin(\omega t)\). Energy ratios can be quickly found using \(PE:KE = x^2 : (A^2 - x^2)\).

Answer 1

The Correct Option is B

Concept: For SHM starting from the mean position, displacement is \[ x = A\sin(\omega t) \] Potential Energy: \[ PE = \frac{1}{2}k x^2 \] Kinetic Energy: \[ KE = \frac{1}{2}k(A^2 - x^2) \] Thus, \[ \frac{PE}{KE} = \frac{x^2}{A^2 - x^2} \]

Step 1: Substitute \(t=\dfrac{T}{6}\). Angular frequency relation: \[ \omega = \frac{2\pi}{T} \] Thus, \[ \omega t = \frac{2\pi}{T} \times \frac{T}{6} \] \[ \omega t = \frac{\pi}{3} \]

Step 2: Find displacement. \[ x = A\sin\left(\frac{\pi}{3}\right) \] \[ x = A\left(\frac{\sqrt{3}}{2}\right) \] \[ x^2 = \frac{3A^2}{4} \]

Step 3: Compute energy ratio. \[ \frac{PE}{KE} = \frac{\frac{3A^2}{4}}{A^2 - \frac{3A^2}{4}} \] \[ = \frac{\frac{3A^2}{4}}{\frac{A^2}{4}} \] \[ = 3 \] Thus, \[ PE : KE = 3 : 1 \]