Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20 \text{ m/s}^2$ at a distance of $5 \text{ m}$ from the mean position. The time period of oscillation is:

$2\pi \text{ s}$

$\pi \text{ s}$

$2 \text{ s}$

$1 \text{ s}$

Step-by-Step Solution

Acceleration in SHM: For a particle (or pendulum bob behaving as a simple harmonic oscillator) executing Simple Harmonic Motion, the magnitude of acceleration $a$ is related to the displacement $x$ (distance from the mean position) by the formula: $a = \omega^2 x$ where $\omega$ is the angular frequency.
Given Data: Acceleration $a = 20 \text{ m/s}^2$ Displacement $x = 5 \text{ m}$
Calculate Angular Frequency ( $\omega$ ): Substituting the values into the formula: $20 = \omega^2 (5)$ $\omega^2 = \frac{20}{5} = 4$ $\omega = \sqrt{4} = 2 \text{ rad/s}$
Calculate Time Period ( $T$ ): The time period $T$ is given by the relation: $T = \frac{2\pi}{\omega}$ Substitute $\omega = 2$ : $T = \frac{2\pi}{2} = \pi \text{ s}$

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