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

Step 1: Understanding the Question:
The particle performs Simple Harmonic Motion (SHM) starting from the mean position.
We need to find the energy distribution (Potential vs. Kinetic) at a specific point in time.
Step 2: Key Formula or Approach:
1. Displacement from mean position: \( x = A \sin(\omega t) \).
2. Angular frequency: \( \omega = \frac{2\pi}{T} \).
3. Potential Energy (P.E.): \( U = \frac{1}{2} k x^2 \).
4. Kinetic Energy (K.E.): \( K = \frac{1}{2} k (A^2 - x^2) \).
5. Ratio: \( \frac{U}{K} = \frac{x^2}{A^2 - x^2} \).
Step 3: Detailed Explanation:
Calculate the phase angle at \( t = T/6 \):
\[ \omega t = \left( \frac{2\pi}{T} \right) \left( \frac{T}{6} \right) = \frac{\pi}{3} = 60^\circ \]
Find the displacement \(x\) at this time:
\[ x = A \sin(60^\circ) = A \left( \frac{\sqrt{3}}{2} \right) \]
Squaring the displacement:
\[ x^2 = A^2 \left( \frac{3}{4} \right) \]
Now, find the ratio of P.E. to K.E.:
\[ \frac{U}{K} = \frac{x^2}{A^2 - x^2} \]
Substitute \( x^2 = \frac{3}{4}A^2 \):
\[ \frac{U}{K} = \frac{\frac{3}{4}A^2}{A^2 - \frac{3}{4}A^2} = \frac{\frac{3}{4}A^2}{\frac{1}{4}A^2} \]
\[ \frac{U}{K} = 3 \]
Step 4: Final Answer:
The ratio of potential energy to kinetic energy is \(3:1\).