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 Simple Harmonic Motion (SHM) is the sum of kinetic energy (K.E.) and potential energy (P.E.), both dependent on displacement \( x \). Let \( A \) represent the amplitude. If the displacement is \( x = \frac{A}{n} \), then the potential energy is \( P.E. = \frac{1}{2} k x^2 \), and the kinetic energy is \( K.E. = E - P.E. = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 \). This simplifies to \( K.E. = \frac{1}{2} k (A^2 - x^2) \). Substituting \( x = \frac{A}{n} \) yields \( P.E. = \frac{1}{2} k \left(\frac{A}{n}\right)^2 = \frac{1}{2} k \frac{A^2}{n^2} \) and \( K.E. = \frac{1}{2} k \left(A^2 - \frac{A^2}{n^2}\right) = \frac{1}{2} k A^2 \left(1 - \frac{1}{n^2}\right) \). The ratio of kinetic energy to potential energy is \( \frac{K.E.}{P.E.} = \frac{1 - \frac{1}{n^2}}{\frac{1}{n^2}} = n^2 - 1:1 \).