Analyzing the Oscillation Frequency of Kinetic Energy in Simple Harmonic Motion

This question delves into the specifics of Simple Harmonic Motion (SHM), asking how the frequency of a particle's kinetic energy relates to its primary oscillation frequency. We are given the displacement formula $x(t) = A \cos(\omega t + \phi)$ and the oscillation frequency $f$. We need to find the frequency at which the kinetic energy oscillates.

Understanding Simple Harmonic Motion (SHM)

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. The displacement of a particle in SHM can be represented as:

$ x(t) = A \cos(\omega t + \phi) $

• A represents the amplitude (maximum displacement).
• \\omega represents the angular frequency.
• t represents time.
• \\phi represents the phase constant.

The frequency ($f$) of oscillation is related to the angular frequency ($\omega$) by the formula:

$ f = \frac{\omega}{2\pi} $

Deriving Kinetic Energy in SHM

To find the frequency of the kinetic energy, we first need to determine the expression for kinetic energy ($KE$). Kinetic energy is given by the formula:

$ KE = \frac{1}{2}mv^2 $

where $m$ is the mass and $v$ is the velocity of the particle.

First, let's find the velocity ($v(t)$) by differentiating the displacement ($x(t)$) with respect to time ($t$):

$ v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \cos(\omega t + \phi)] $

$ v(t) = -A\omega \sin(\omega t + \phi) $

Now, substitute this velocity expression into the kinetic energy formula:

$ KE(t) = \frac{1}{2}m[-A\omega \sin(\omega t + \phi)]^2 $

$ KE(t) = \frac{1}{2}m A^2 \omega^2 \sin^2(\omega t + \phi) $

Analyzing the Frequency of Kinetic Energy Oscillation

The expression for kinetic energy involves $\sin^2(\omega t + \phi)$. To find the frequency of oscillation, we need to simplify this term. We use the trigonometric identity:

$ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} $

Applying this identity with $\theta = \omega t + \phi$:

$ KE(t) = \frac{1}{2}m A^2 \omega^2 \left( \frac{1 - \cos(2(\omega t + \phi))}{2} \right) $

$ KE(t) = \frac{1}{4}m A^2 \omega^2 (1 - \cos(2\omega t + 2\phi)) $

$ KE(t) = \left( \frac{1}{4}m A^2 \omega^2 \right) - \left( \frac{1}{4}m A^2 \omega^2 \right) \cos(2\omega t + 2\phi) $

This expression shows that the kinetic energy has two components:

• A constant term: $\frac{1}{4}m A^2 \omega^2$
• An oscillating term: $-\frac{1}{4}m A^2 \omega^2 \cos(2\omega t + 2\phi)$

The oscillating term depends on $\cos(2\omega t + 2\phi)$. The angular frequency of this oscillation is $2\omega$.

Relating Kinetic Energy Frequency to Particle Frequency

We know the angular frequency of the particle's SHM is $\omega$, and its frequency is $f = \frac{\omega}{2\pi}$.

The angular frequency of the kinetic energy oscillation is $2\omega$. Let the frequency of the kinetic energy oscillation be $f_{KE}$. Then:

$ f_{KE} = \frac{\text{Angular frequency of KE}}{2\pi} = \frac{2\omega}{2\pi} $

$ f_{KE} = 2 \left( \frac{\omega}{2\pi} \right) $

Since $f = \frac{\omega}{2\pi}$, we can substitute $f$ into the equation:

$ f_{KE} = 2f $

Therefore, the kinetic energy oscillates with twice the frequency of the particle's simple harmonic motion.

Conclusion on Kinetic Energy Frequency

The kinetic energy of a particle in SHM oscillates at a frequency that is double the frequency of the particle's displacement oscillation. If the particle's oscillation frequency is $f$, its kinetic energy oscillates with a frequency of $2f$.