Question

Find the ratio of K.E. and P.E. when a particle performs SHM when it is at \( \frac{1}{n} \) times of its amplitude from the mean position.

• A. \( \frac{n^2}{2} \)
• B. \( n^2 + 1 \)
• C. \( n^2 - 1 \)
• D. \( n^2 \)

Hint:

In SHM, the total energy is conserved, and the sum of K.E. and P.E. is constant. Use the relations for \( K.E. \) and \( P.E. \) at specific positions to compute their ratios.

Answer 1

The Correct Option is C

Step 1: Potential Energy and Kinetic Energy in SHM.
The total energy in SHM is: \[ E_{\text{total}} = K.E. + P.E. \] Let the amplitude of the particle be \( A \), and the position at any instant is \( x \). The potential energy is: \[ P.E. = \frac{1}{2} k x^2, \] where \( k \) is the spring constant. The kinetic energy is: \[ K.E. = \frac{1}{2} k (A^2 - x^2). \] Step 2: At \( x = \frac{A}{n} \).
Substitute \( x = \frac{A}{n} \): \[ P.E. = \frac{1}{2} k \left( \frac{A}{n} \right)^2 = \frac{1}{2} k \frac{A^2}{n^2}. \] \[ K.E. = \frac{1}{2} k \left( A^2 - \left( \frac{A}{n} \right)^2 \right) = \frac{1}{2} k \left( A^2 - \frac{A^2}{n^2} \right). \] Simplify: \[ K.E. = \frac{1}{2} k A^2 \left( 1 - \frac{1}{n^2} \right) = \frac{1}{2} k A^2 \frac{n^2 - 1}{n^2}. \] Step 3: Ratio of \( K.E. \) to \( P.E. \).
The ratio is: \[ \frac{K.E.}{P.E.} = \frac{\frac{1}{2} k A^2 \frac{n^2 - 1}{n^2}}{\frac{1}{2} k \frac{A^2}{n^2}}. \] Simplify: \[ \frac{K.E.}{P.E.} = \frac{n^2 - 1}{1}. \] Thus: \[ \frac{K.E.}{P.E.} = n^2 - 1. \] Step 4: Final Answer.
The ratio of \( K.E. \) to \( P.E. \) is: \[ \boxed{n^2 - 1}. \]