Question

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

• A. 1:4
• B. 1:3
• C. 2:1
• D. 1:1

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between potential energy (PE) and kinetic energy (KE) in Simple Harmonic Motion (SHM). The total mechanical energy \( E \) in SHM is constant and can be expressed as the sum of potential energy and kinetic energy:

E = PE + KE

The potential energy for a particle executing SHM at a displacement \( x \) can be expressed as:

PE = \frac{1}{2} k x^2

Here, \( k \) is the force constant, and \( x \) is the displacement from the mean position.

The kinetic energy is given by:

KE = \frac{1}{2} k (A^2 - x^2)

Here, \( A \) is the amplitude of the motion.

When the displacement \( x \) is half of the amplitude \( A \), i.e., \( x = \frac{A}{2} \), the potential and kinetic energies become:

Substituting \( x = \frac{A}{2} \) into the equation for potential energy:

PE = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \frac{A^2}{4} = \frac{1}{8} k A^2

And substituting \( x = \frac{A}{2} \) into the equation for kinetic energy:

KE = \frac{1}{2} k \left(A^2 - \left(\frac{A}{2}\right)^2\right) = \frac{1}{2} k \left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2} k \frac{3A^2}{4} = \frac{3}{8} k A^2

Now, we can find the ratio of potential energy to kinetic energy:

\text{Ratio} = \frac{PE}{KE} = \frac{\frac{1}{8} k A^2}{\frac{3}{8} k A^2} = \frac{1}{3}

Therefore, the ratio of potential energy to kinetic energy when the displacement is half the amplitude is 1:3.