Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

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

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

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

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

Step-by-Step Solution

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.
Substitute Given Values:

Amplitude $A = a$
Displacement $x = \frac{a}{2}$
Angular frequency $\omega = \frac{2\pi}{T}$

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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