Question

A particle starts oscillating simple harmonically from its mean position with time period 'T'. At time $t = T/6$, the ratio of the potential energy to kinetic energy of the particle is \dots

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

Hint:

A useful shortcut for SHM energy ratios: Since $PE \propto \sin^2(\theta)$ and $KE \propto \cos^2(\theta)$, the ratio $PE/KE$ is simply $\tan^2(\theta)$. At $t=T/6$, $\theta = 60^\circ$. $\tan^2(60^\circ) = (\sqrt{3})^2 = 3$. Ratio is 3:1!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a particle starting from the mean position, displacement $x = A \sin(\omega t)$, where $\omega = 2\pi/T$.

Step 2: Formula Application:
At $t = T/6$: $x = A \sin \left( \frac{2\pi}{T} \cdot \frac{T}{6} \right) = A \sin(\pi/3) = A \frac{\sqrt{3}}{2}$. Potential Energy $P.E. = \frac{1}{2} kx^2$ and Kinetic Energy $K.E. = \frac{1}{2} k(A^2 - x^2)$.

Step 3: Explanation:
$P.E. = \frac{1}{2} k \left( \frac{3A^2}{4} \right)$. $K.E. = \frac{1}{2} k \left( A^2 - \frac{3A^2}{4} \right) = \frac{1}{2} k \left( \frac{A^2}{4} \right)$. Ratio $P.E. / K.E. = \frac{3A^2/4}{A^2/4} = 3 / 1$.

Step 4: Final Answer:
The ratio of potential energy to kinetic energy is 3 : 1.