Question

A simple pendulum starts oscillating simple harmonically from its mean position (\(x = 0\)) with amplitude ' \(a\) ' and periodic time ' \(T\) '. The magnitude of velocity of pendulum at \(x = \frac{a}{2}\) is

• A. \(\frac{3\pi^2 a}{T}\)
• B. \(\frac{\sqrt{3}\pi a}{2T}\)
• C. \(\frac{\pi a}{T}\)
• D. \(\frac{\sqrt{3}\pi a}{T}\)

Hint:

Maximum velocity $v_{max} = \omega a$ occurs at the mean position ($x=0$), while velocity is zero at the extreme positions ($x=a$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the instantaneous velocity of a particle in SHM at a given displacement.
Step 2: Key Formula or Approach:
1) Velocity in SHM: \(v = \omega \sqrt{a^2 - x^2}\).
2) Angular frequency: \(\omega = \frac{2\pi}{T}\).
Step 3: Detailed Explanation:
Given displacement \(x = a/2\).
Substitute \(x\) into the velocity formula:
\[ v = \omega \sqrt{a^2 - \left( \frac{a}{2} \right)^2} \]
\[ v = \omega \sqrt{a^2 - \frac{a^2}{4}} \]
\[ v = \omega \sqrt{\frac{3a^2}{4}} \]
\[ v = \omega \cdot \frac{\sqrt{3}a}{2} \]
Now, substitute \(\omega = \frac{2\pi}{T}\):
\[ v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}a}{2} \]
\[ v = \frac{\sqrt{3}\pi a}{T} \]
Step 4: Final Answer:
The magnitude of velocity is \(\frac{\sqrt{3}\pi a}{T}\).