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: Velocity Formula
The velocity of a particle in SHM is given by $v = \omega \sqrt{a^2 - x^2}$.

Step 2: Substitution
Substitute $x = a/2$ and $\omega = 2\pi/T$:
$v = \frac{2\pi}{T} \sqrt{a^2 - (a/2)^2}$.

Step 3: Calculation
$v = \frac{2\pi}{T} \sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \sqrt{\frac{3a^2}{4}}$.
$v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}a}{2} = \frac{\sqrt{3}\pi a}{T}$.
Final Answer: (D)