A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The

Question

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x = rac{a}{2}$ will be:

Accepted Answer

1. **Identify the Speed Equation in SHM:** The speed $v$ of a particle executing simple harmonic motion at a displacement $x$ from the mean position is given by $v = \omega \sqrt{A^2 - x^2}$, where $A$ is the amplitude and $\omega$ is the angular frequency.
2. **Substitute Given Values:** 
   - Amplitude $A = a$
   - Displacement $x = \frac{a}{2}$
   - Angular frequency $\omega = \frac{2\pi}{T}$
3. **Calculate Speed:**
   $$v = \frac{2\pi}{T} \sqrt{a^2 - \left(\frac{a}{2}\right)^2}$$
   $$v = \frac{2\pi}{T} \sqrt{a^2 - \frac{a^2}{4}}$$
   $$v = \frac{2\pi}{T} \sqrt{\frac{3a^2}{4}}$$
   $$v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T}$$

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