Animated Solution for Physics - Oscillations: A particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $\nu$. The average kinetic energy during its motion from the position of equilibrium to the end is

The Dance of SHM

Imagine a particle of mass $m$ executing simple harmonic motion (SHM). As it glides from the equilibrium position ($x=0$) to the extreme end ($x=a$), its velocity gracefully decreases from a maximum value to absolute zero.

This journey from the center to the edge represents exactly one-fourth of its full oscillation cycle. If the total time period is $T$, this specific motion takes a time interval of $T/4$. To find the average kinetic energy during this phase, we must look at how the kinetic energy varies with time.

The kinetic energy $K$ of a particle in SHM at any instant $t$ is given by the equation: where $\omega$ is the angular frequency and $a$ is the amplitude.

To find the time-averaged kinetic energy over the interval from $t=0$ to $t=T/4$, we set up the integral:

Now, we could evaluate this integral the long way, but there is a beautiful mathematical shortcut that every physics student should know.

The average value of ${\cos }^2(\theta )$ or ${\sin }^2(\theta )$ over any quarter cycle, half cycle, or full cycle is always exactly $\frac{1}{2}$. This is because the area under the squared sine and cosine curves are perfectly symmetric and sum to 1 over a cycle.

Using this powerful shortcut, we can instantly write:

Substituting this back into our kinetic energy expression, the average kinetic energy becomes:

Notice how the maximum kinetic energy ($\frac{1}{2}m{\omega }^2{a}^2$) is simply halved!

Frequency Substitution

We have our answer, but there is a slight catch. If you look at the options, they are given in terms of the linear frequency $ u$, not the angular frequency $\omega$.

We know the fundamental relationship connecting them:

Let's substitute this into our average kinetic energy equation:

Squaring the term inside the bracket gives $4\pi^2 u^2$. The $4$ in the numerator perfectly cancels the $4$ in the denominator:

And there we have it! The final average kinetic energy is $\pi^2 m a^2 u^2$.

Before we conclude, let's discuss a classic trap. What if the question had asked for the average kinetic energy over distance (or space) instead of time?

In that case, we would integrate with respect to $x$:

The space average ($\frac{1}{3}$) is different from the time average ($\frac{1}{4}$). Unless explicitly stated otherwise, "average" in kinematics always implies the time average. Always read the question carefully!