Question

The amplitude of a particle executing simple harmonic motion is 6 cm. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes 4:5 is

• A. 6 cm
• B. 4 cm
• C. 3 cm
• D. 2 cm

Hint:

If \( \frac{U}{K} = \frac{m}{n} \), then \( \frac{x^2}{A^2 - x^2} = \frac{m}{n} \implies x^2 = \frac{m}{m+n} A^2 \implies x = A \sqrt{\frac{m}{m+n}} \). This is a faster shortcut.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the total energy is the sum of kinetic energy (KE) and potential energy (PE). Both energies vary with position. We are given the ratio PE:KE at a specific distance \( x \) and need to find \( x \).
Step 2: Key Formula or Approach:
1. Potential Energy: \( U = \frac{1}{2} k x^2 \) 2. Kinetic Energy: \( K = \frac{1}{2} k (A^2 - x^2) \) 3. Given ratio: \( \frac{U}{K} = \frac{4}{5} \)
Step 3: Detailed Explanation:
Let the amplitude be \( A = 6 \) cm. Let the displacement be \( x \). We are given: \[ \frac{U}{K} = \frac{4}{5} \] Substitute the formulas: \[ \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{4}{5} \] Cancel common terms (\( \frac{1}{2} k \)): \[ \frac{x^2}{A^2 - x^2} = \frac{4}{5} \] Cross-multiply: \[ 5x^2 = 4(A^2 - x^2) \] \[ 5x^2 = 4A^2 - 4x^2 \] \[ 9x^2 = 4A^2 \] Take square root of both sides: \[ 3x = 2A \] \[ x = \frac{2}{3} A \] Substitute \( A = 6 \) cm: \[ x = \frac{2}{3} (6) = 4 \, \text{cm} \]
Step 4: Final Answer:
The distance is 4 cm.