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: Total Energy in SHM. The total energy \( E_{\text{total}} \) in simple harmonic motion (SHM) is the sum of kinetic energy (K.E.) and potential energy (P.E.):
\[E_{\text{total}} = K.E. + P.E.\]Given the amplitude \( A \) and instantaneous position \( x \), the potential energy is defined as:
\[P.E. = \frac{1}{2} k x^2,\]where \( k \) represents the spring constant. The kinetic energy is given by:
\[K.E. = \frac{1}{2} k (A^2 - x^2).\]Step 2: Energy at Position \( x = \frac{A}{n} \).
When the position is \( x = \frac{A}{n} \), the potential energy becomes:
\[P.E. = \frac{1}{2} k \left( \frac{A}{n} \right)^2 = \frac{1}{2} k \frac{A^2}{n^2}.\]The kinetic energy at this position is:
\[K.E. = \frac{1}{2} k \left( A^2 - \left( \frac{A}{n} \right)^2 \right) = \frac{1}{2} k \left( A^2 - \frac{A^2}{n^2} \right).\]Simplifying the kinetic energy expression yields:
\[K.E. = \frac{1}{2} k A^2 \left( 1 - \frac{1}{n^2} \right) = \frac{1}{2} k A^2 \frac{n^2 - 1}{n^2}.\]Step 3: Ratio of Kinetic Energy to Potential Energy.
The ratio of kinetic energy to potential energy is calculated as:
\[\frac{K.E.}{P.E.} = \frac{\frac{1}{2} k A^2 \frac{n^2 - 1}{n^2}}{\frac{1}{2} k \frac{A^2}{n^2}}.\]Upon simplification:
\[\frac{K.E.}{P.E.} = \frac{n^2 - 1}{1}.\]This simplifies to:
\[\frac{K.E.}{P.E.} = n^2 - 1.\]Step 4: Conclusion.
The ratio of kinetic energy to potential energy is determined to be:
\[\boxed{n^2 - 1}.\]