Rankers Physics

Energy in SHM: Practice Problem & Solution

The particle executing simple harmonic motion has a kinetic energy $K_0 \cos^2 \omega t$. The maximum values of the potential energy and the total energy are respectively: (2007)
$K_0/2 \text{ and } K_0$
$K_0 \text{ and } 2K_0$
$K_0 \text{ and } K_0$
$0 \text{ and } 2K_0$

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 maximum kinetic energy is $K_0$.\nIn an ideal SHM without damping, the total energy remains conserved and equals the maximum kinetic energy, which is $K_0$.\nThe maximum potential energy is also equal to the total energy, which is $K_0$.

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Similar Practice Questions

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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
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3. Assertion (A): A SHM may be assumed as composition of many SHM's. Reason (R): Superposition of many SHM's (along same line) of same frequency will be a SHM.  
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