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

Question

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

Accepted Answer

1. **Identify Formulas:** The speed $v$ of a particle in SHM is given by $v = \omega \sqrt{A^2 - x^2}$, where $A$ is the amplitude and $x$ is the displacement . The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$ .
2. **Substitute Values:**
   - Amplitude $A = a$
   - Displacement $x = \frac{a}{2}$
   - $\omega = \frac{2\pi}{T}$
3. **Calculate Speed:**
   $$ v = \frac{2\pi}{T} \sqrt{a^2 - \left(\frac{a}{2}\right)^2} $$
   $$ v = \frac{2\pi}{T} \sqrt{a^2 - \frac{a^2}{4}} $$
   $$ v = \frac{2\pi}{T} \sqrt{\frac{3a^2}{4}} $$
   $$ v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2} $$
   $$ v = \frac{\pi a\sqrt{3}}{T} $$

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