Time Period of Simple Harmonic Motion – Rankers Physics
Topic: Oscillation
Subtopic: Equation of SHM

Time Period of Simple Harmonic Motion

A particle executes SHM according to equation \(2\frac{d^2x}{dt^2} + 100x = 0\) (where \(x\) is in m and \(t\) is in second). Its time period of oscillation is
\(\frac{5\pi}{\sqrt{2}}\text{ s}\)
\(5\sqrt{2}\pi\text{ s}\)
\(2\sqrt{5}\pi\text{ s}\)
\(\frac{\sqrt{2}}{5}\pi\text{ s}\)

Solution:

Divide the given equation by 2 to get \(\frac{d^2x}{dt^2} + 50x = 0\). Comparing this with the standard SHM differential equation \(\frac{d^2x}{dt^2} + \omega^2x = 0\) gives \(\omega = \sqrt{50} = 5\sqrt{2}\text{ rad/s}\). The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).

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