Animated Solution for Physics - Oscillations: Starting from the origin, a body oscillates simple harmonically with a period of $2\text{ s}$. After what time will its kinetic energy be $75\%$ of the total energy?

Imagine a body oscillating back and forth, starting right from the mean position, the origin. As it moves, its energy constantly dances between kinetic and potential forms.

Because the body starts from the origin, its displacement is perfectly described by a sine function:

This means its velocity, which is the rate of change of displacement, starts at a maximum and follows a cosine curve:

Consequently, its kinetic energy, which depends on the square of the velocity, starts at a maximum and follows a cosine squared curve:

The question asks for the exact moment when the kinetic energy drops to $75\%$ of the total energy $E$. Let's set up our equation by equating our kinetic energy formula to three-fourths of $E$:

We can elegantly cancel the total energy $E$ from both sides. Taking the square root, we find that the cosine of $\omega t$ must be equal to $\frac{\sqrt{3}}{2}$. We take the positive root because we are looking for the first time this energy drop happens after $t=0$.

Now, what angle gives a cosine of $\frac{\sqrt{3}}{2}$? That's right, $30^{\circ }$, or $\frac{\pi }{6}$ radians. So, we can write:

We know the angular frequency $\omega$ is related to the time period $T$ by the formula $\omega =\frac{2\pi }{T}$. Let's substitute that into our equation:

The time period $T$ is given as $2\text{ s}$. Plugging that in, the $2\pi$ terms cancel out beautifully:

And there we have it! Exactly one-sixth of a second after starting from the mean position, the kinetic energy of the oscillator drops to $75\%$ of its total energy.