Question

A particle starts oscillating simple harmonically from its mean position with time period \(T\); find the ratio of potential energy to kinetic energy at time \(t = \frac{T}{6}\).

• A. \( \frac{1}{3} \)
• B. \( 3 \)
• C. \( \frac{1}{2} \)
• D. \( 1 \)

Hint:

In SHM starting from the mean position, use \(x = A\sin(\omega t)\). The ratio \( \frac{U}{K} \) can quickly be found using \[ \frac{U}{K} = \frac{x^2}{A^2 - x^2}. \]

Answer 1

The Correct Option is B

Concept: In Simple Harmonic Motion (SHM):

• Displacement: \(x = A \sin(\omega t)\) (when motion starts from mean position)
• Angular frequency: \( \omega = \frac{2\pi}{T} \)
• Potential Energy: \( U = \frac{1}{2}k x^2 \)
• Total Energy: \( E = \frac{1}{2}kA^2 \)
• Kinetic Energy: \( K = E - U \)

Thus, \[ \frac{U}{E} = \frac{x^2}{A^2} \] and \[ \frac{U}{K} = \frac{x^2}{A^2 - x^2} \]
Step 1: {Find the displacement at \(t = \frac{T}{6}\).} Since \[ x = A\sin(\omega t) \] and \[ \omega = \frac{2\pi}{T} \] \[ \omega t = \frac{2\pi}{T} \cdot \frac{T}{6} \] \[ \omega t = \frac{\pi}{3} \] Thus, \[ x = A\sin\left(\frac{\pi}{3}\right) \] \[ x = A\left(\frac{\sqrt{3}}{2}\right) \]
Step 2: {Find potential energy fraction.} \[ \frac{x^2}{A^2} = \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ \frac{x^2}{A^2} = \frac{3}{4} \] Thus, \[ U = \frac{3}{4}E \]
Step 3: {Find kinetic energy.} \[ K = E - U \] \[ K = E - \frac{3}{4}E \] \[ K = \frac{1}{4}E \]
Step 4: {Find the required ratio.} \[ \frac{U}{K} = \frac{\frac{3}{4}E}{\frac{1}{4}E} \] \[ \frac{U}{K} = 3 \]