Question

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

Hint:

At \(t = T/12\), \(x = A/2\), \(PE/KE = 1/3\). At \(t = T/8\), \(x = A/\sqrt{2}\), \(PE = KE\). At \(t = T/6\), \(x = A\sqrt{3}/2\), \(PE/KE = 3/1\). Memorizing these standard points saves valuable time!

Answer 1

Step 1: Understanding the Question:
The displacement from mean position is \( x = A \sin(\omega t) \). We need PE (\( \frac{1}{2}kx^2 \)) and KE (\( \frac{1}{2}k(A^2-x^2) \)).
Step 2: Key Formula or Approach:
Ratio \( \frac{PE}{KE} = \frac{x^2}{A^2 - x^2} \).
Step 3: Detailed Explanation:
At \( t = T/6 \), phase \( \theta = \omega t = \frac{2\pi}{T} \cdot \frac{T}{6} = \frac{\pi}{3} \).
\( x = A \sin(\pi/3) = \frac{\sqrt{3}A}{2} \).
Ratio:
\[ \frac{PE}{KE} = \frac{(\sqrt{3}A/2)^2}{A^2 - (\sqrt{3}A/2)^2} = \frac{3A^2/4}{A^2/4} = 3 \]
Step 4: Final Answer:
The ratio is \( 3:1 \).