Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\frac{\pi a\sqrt{3}}{2T}$

$\frac{\pi a}{T}$

$\frac{3\pi a}{T}$

$\frac{\pi a\sqrt{3}}{T}$

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 $\omega = \frac{2\pi}{T}$ .
Substitute Values:

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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