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 Question:
A particle undergoes Simple Harmonic Motion (SHM) starting from the mean position. We must find the ratio of its Potential Energy (PE) to its Kinetic Energy (KE) at a specific time snapshot $t = T/6$.

Step 2: Key Formula or Approach:
1. Displacement starting from mean position: $x(t) = A \sin(\omega t) = A \sin\left(\frac{2\pi}{T} t\right)$.
2. Total Energy (TE) in SHM: $E = \frac{1}{2} k A^2$.
3. Potential Energy (PE): $PE = \frac{1}{2} k x^2$.
4. Kinetic Energy (KE): $KE = TE - PE$.

Step 3: Detailed Explanation:
First, find the displacement $x$ at $t = T/6$:
$$x = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{6}\right)$$
$$x = A \sin\left(\frac{\pi}{3}\right) = A \sin(60^\circ)$$
Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, we have $x = A\frac{\sqrt{3}}{2}$.
Now, calculate the Potential Energy (PE) at this position:
$$PE = \frac{1}{2} k \left(A\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} k \left( A^2 \frac{3}{4} \right) = \frac{3}{4} \left(\frac{1}{2} k A^2\right)$$
$$PE = \frac{3}{4} E$$
Since Total Energy is conserved ($PE + KE = E$), calculate the Kinetic Energy (KE):
$$KE = E - PE = E - \frac{3}{4}E = \frac{1}{4} E$$
Finally, find the required ratio of PE to KE:
$$\text{Ratio} = \frac{PE}{KE} = \frac{\frac{3}{4} E}{\frac{1}{4} E} = \frac{3}{1}$$

Step 4: Final Answer:
The ratio is 3 : 1, matching option (d).