Angular SHM and Simple Pendulum: Practice Problem & Solution
A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration √3g m / s² .The period of small oscillations of the pendulum about its equilibrium position is (g = π² m/s²):
Solution Explained:
To solve this problem, we apply the core principles of Angular SHM and Simple Pendulum. Understanding the underlying formula is key to arriving at the correct answer below:
\[ T = 2\pi\sqrt{\frac{l}{\sqrt{a^{2}+g^{2}}}} \]
\[ T = 2\pi\sqrt{\frac{0.5}{\sqrt{\left( \sqrt{3}g \right)^{2}+g^{2}}}} \]
\[ T = 2\pi\sqrt{\frac{0.5}{2g}} \]
\[ T = 2\pi\frac{1}{2\sqrt{g}} = 2\pi\frac{1}{2\pi}= 1 sec \]
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