Question

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

• A. \( n^2 - 1:1 \)
• B. \( 1:n^2 - 1 \)
• C. \( 1:n \)
• D. \( n:1 \)

Hint:

The kinetic and potential energy in SHM are always complementary and their sum remains constant.

Answer 1

The Correct Option is A

The total energy in SHM is given by: \[ E = K.E. + P.E., \] where \( K.E. \) (kinetic energy) and \( P.E. \) (potential energy) are related to the displacement \( x \). 

Let \( A \) be the amplitude, and \( x = \frac{A}{n} \). Then: \[ P.E. = \frac{1}{2} k x^2, \quad K.E. = E - P.E. = \frac{1}{2} k A^2 - \frac{1}{2} k x^2. \] Simplify: \[ K.E. = \frac{1}{2} k (A^2 - x^2). \] 

Substitute \( x = \frac{A}{n} \): \[ P.E. = \frac{1}{2} k \left(\frac{A}{n}\right)^2 = \frac{1}{2} k \frac{A^2}{n^2}, \] \[ K.E. = \frac{1}{2} k \left(A^2 - \frac{A^2}{n^2}\right) = \frac{1}{2} k A^2 \left(1 - \frac{1}{n^2}\right). \] 

The ratio \( \frac{K.E.}{P.E.} \) is: \[ \frac{K.E.}{P.E.} = \frac{1 - \frac{1}{n^2}}{\frac{1}{n^2}} = n^2 - 1:1. \]