Energy in SHM

Practice Questions:

Question 11: easy

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:

1. \(\frac{\pi}{100}\text{ s}\)

2. \(\frac{\pi}{50}\text{ s}\)

3. \(\frac{\pi}{20}\text{ s}\)

4. \(\frac{\pi}{10}\text{ s}\)

View Answer

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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Question 12: easy

A particle is performing simple harmonic motion of amplitude \(A\) about origin. Then position at which kinetic energy of particle is 8 times of its potential energy at that instant is:

1. \(x = \frac{A}{2\sqrt{2}}\)

2. \(x = \frac{A}{8}\)

3. \(x = \frac{A}{3}\)

4. \(x = \frac{A}{9}\)

View Answer

The given condition is \(\text{KE} = 8\text{PE}\). Substituting the expressions: \(\frac{1}{2}m\omega^2(A^2 - x^2) = 8\left(\frac{1}{2}m\omega^2 x^2\right)\), which simplifies to \(A^2 - x^2 = 8x^2\) or \(9x^2 = A^2\). Thus, \(x = \frac{A}{3}\).

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