Rankers Physics

Angular SHM and Simple Pendulum: Practice Problem & Solution

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $ T = 2\pi \sqrt{(l/g')} $, where $ g' $ is equal to: (1991)
g
g - a
g + a
$ \sqrt{(g^2 + a^2)} $

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:

For a trolley accelerating horizontally, the effective gravity is the vector sum of standard gravity $ g $ (downwards) and the pseudo acceleration $ a $ (backwards). Hence, $ g' = \sqrt{g^2 + a^2} $.

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