Question

When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is:

• A. 1 : 3
• B. 2 : 1
• C. 3 : 1
• D. 1 : 2
• E. 2 : 3

Hint:

At $y = A/2$, $PE$ is $25%$ of total energy and $KE$ is $75%$ of total energy. Thus the ratio is $75:25 = 3:1$.

Answer 1

The Correct Option is C

To solve this problem, we need to find the ratio of the kinetic energy to potential energy of a particle executing simple harmonic motion (SHM) when its displacement is half its amplitude.

Let's consider the following variables:

• Amplitude of the SHM: \(A\)
• Displacement: \(x = \frac{A}{2}\)

The potential energy (PE) in SHM when displacement is \(x\) is given by:

\(\text{PE} = \frac{1}{2} k x^2\)

where \(k\) is the spring constant.

The kinetic energy (KE) is given by:

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

Substituting \(x = \frac{A}{2}\) into the equations:

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

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

Now, we find the ratio of KE to PE:

\(\frac{\text{KE}}{\text{PE}} = \frac{\frac{3}{8} k A^2}{\frac{1}{8} k A^2} = \frac{3}{1} = 3:1\)

Therefore, the correct answer is 3 : 1.