Equation of SHM: Practice Problem & Solution
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: (2018)
Solution Explained:
To solve this problem, we apply the core principles of Equation of SHM. Understanding the underlying formula is key to arriving at the correct answer below:
Acceleration $a = \omega^2 x \Rightarrow 20 = \omega^2 (5) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \text{ rad/s}$. Time period $T = \frac{2\pi}{\omega} = \pi \text{ s}$.
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