Question

For a particle executing simple harmonic motion, the ratio of kinetic and potential energies at a point where displacement is one half of the amplitude is

• A. \(3 : 1\)
• B. \(1 : 3\)
• C. \(2 : 1\)
• D. \(1 : 2\)

Hint:

At displacement \(x = \dfrac{A}{2}\) in SHM, kinetic and potential energies follow the ratio \(K:U = 3:1\), derived from energy conservation.

Answer 1

The Correct Option is A

Step 1: Use energy relations in SHM 
In simple harmonic motion, the total mechanical energy \(E = \frac{1}{2}kA^2\) is constant. At any displacement \(x\), the potential energy \(U = \frac{1}{2}kx^2\), and kinetic energy \(K = E - U = \frac{1}{2}k(A^2 - x^2)\). 
Step 2: Use the given displacement 
Given \(x = \dfrac{A}{2}\), so: \[ U = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k\cdot \frac{A^2}{4} = \frac{1}{8}kA^2 \] \[ K = \frac{1}{2}kA^2 - \frac{1}{8}kA^2 = \frac{3}{8}kA^2 \] Step 3: Ratio of Kinetic to Potential energy 
\[ \frac{K}{U} = \frac{\frac{3}{8}kA^2}{\frac{1}{8}kA^2} = \frac{3}{1} \]