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 Concept:
In Simple Harmonic Motion (SHM) starting from the mean position, the displacement is \(x = A \sin(\omega t)\). Potential Energy (P.E.) \(= \frac{1}{2} k x^2\) and Kinetic Energy (K.E.) \(= \text{Total Energy} - \text{P.E.} = \frac{1}{2} k A^2 - \frac{1}{2} k x^2\).

Step 2: Key Formula or Approach:
1. \(x = A \sin \left( \frac{2\pi}{T} t \right)\).
2. \(\frac{PE}{KE} = \frac{x^2}{A^2 - x^2}\).

Step 3: Detailed Explanation:
At \(t = \frac{T}{6}\):
\[ x = A \sin \left( \frac{2\pi}{T} \cdot \frac{T}{6} \right) = A \sin \left( \frac{\pi}{3} \right) \] Since \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\):
\[ x = A \frac{\sqrt{3}}{2} \] Now calculate the squares:
\(x^2 = \frac{3}{4} A^2\).
\(A^2 - x^2 = A^2 - \frac{3}{4} A^2 = \frac{1}{4} A^2\).
Ratio:
\[ \frac{PE}{KE} = \frac{\frac{3}{4} A^2}{\frac{1}{4} A^2} = \frac{3}{1} \]
Step 4: Final Answer:
The ratio of potential energy to kinetic energy is \(3 : 1\).