The potential energy of a harmonic oscillator of mass \(2\text{ kg}\) in its mean position is \(5\text{ J}\). If its total energy is \(9\text{ J}\) and its amplitude is \(0.01\text{ m}\), its time period will be:
\(\frac{\pi}{100}\text{ s}\)
\(\frac{\pi}{50}\text{ s}\)
\(\frac{\pi}{20}\text{ s}\)
\(\frac{\pi}{10}\text{ s}\)
Solution:
The total energy is given by \(E = U_0 + \frac{1}{2}m\omega^2 A^2\). Substituting the values: \(9 = 5 + \frac{1}{2}(2)\omega^2 (0.01)^2\), which gives \(\omega = 200\text{ rad/s}\). Therefore, the time period is \(T = \frac{2\pi}{\omega} = \frac{\pi}{100}\text{ s}\).
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