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

Step 1: Understanding the Question:
The problem requires finding the distribution of energy (Potential vs Kinetic) at a specific time interval in Simple Harmonic Motion.
Step 2: Key Formula or Approach:
Displacement from mean: \( x = A\sin(\omega t) \).
Potential Energy \( PE = \frac{1}{2}kx^2 \).
Kinetic Energy \( KE = \frac{1}{2}k(A^2 - x^2) \).
Ratio \( \frac{PE}{KE} = \frac{x^2}{A^2-x^2} \).
Step 3: Detailed Explanation:
Given \( t = T/6 \) and \( \omega = 2\pi/T \).
Calculate phase angle \( \theta = \omega t \):
\[ \theta = \frac{2\pi}{T} \times \frac{T}{6} = \frac{\pi}{3} = 60^\circ \] Find displacement \(x\):
\[ x = A\sin(60^\circ) = A\left(\frac{\sqrt{3}}{2}\right) \] Square the displacement:
\[ x^2 = \frac{3A^2}{4} \] Calculate energy ratio:
\[ \frac{PE}{KE} = \frac{\frac{3A^2}{4}}{A^2 - \frac{3A^2}{4}} = \frac{\frac{3A^2}{4}}{\frac{A^2}{4}} = \frac{3}{1} \] Step 4: Final Answer:
The ratio of potential energy to kinetic energy is \( 3:1 \).