A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple

Question

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 $20 	ext{ m/s}^2$ at a distance of $5 	ext{ m}$ from the mean position. The time period of oscillation is:

Accepted Answer

1.  **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.
2.  **Given Data:**
    *   Acceleration $a = 20 	ext{ m/s}^2$
    *   Displacement $x = 5 	ext{ m}$
3.  **Calculate Angular Frequency ($\omega$):**
    Substituting the values into the formula:
    $$ 20 = \omega^2 (5) $$
    $$ \omega^2 = \frac{20}{5} = 4 $$
    $$ \omega = \sqrt{4} = 2 	ext{ rad/s} $$
4.  **Calculate Time Period ($T$):**
    The time period $T$ is given by the relation:
    $$ T = \frac{2\pi}{\omega} $$
    Substitute $\omega = 2$:
    $$ T = \frac{2\pi}{2} = \pi 	ext{ s} $$

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