Kinetic and Potential Energy Position in SHM – Rankers Physics

Energy in SHM: Practice Problem & Solution

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:
\(x = \frac{A}{2\sqrt{2}}\)
\(x = \frac{A}{8}\)
\(x = \frac{A}{3}\)
\(x = \frac{A}{9}\)

Solution Explained:

To solve this problem, we apply the core principles of Energy in SHM. Understanding the underlying formula is key to arriving at the correct answer below:

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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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The differential equation for a particle executing S.H.M. is given by \( \frac{d^2 y}{dt^2} + 4y = 0 \), where symbols have their usual meaning. The angular velocity of the particle is given by
2. A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of pendulum at $x = a/2$ will be: (2009)
3. A particle is undergoing SHM having total mechanical energy equal to \(8\text{ J}\). At an instant its kinetic energy is found to be \(10\text{ J}\), then its potential energy at that instant is
4. Assertion (A): In forced oscillations, the steady state motion of the particle (after natural oscillations die out) is SHM whose frequency is the frequency of the driving frequency \(\omega_d\), not the natural frequency \(\omega\) of the particle. Reason (R): In forced oscillation \(\omega_d\) should be greater than natural frequency \(\omega\) of the particle.  
5. Assertion (A): For a particle performing SHM, its speed decreases as it goes away from the mean position. Reason (R): In SHM, the acceleration is always opposite to the velocity of the particle.