Animated Solution for Physics - Oscillations: A particle of mass $1\text{ kg}$ is hanging from a spring of force constant $100{\text{ Nm}}^{-1}$. The mass is pulled slightly downward and released, so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal, is $T/x$. The value of $x$ is.

The beauty of Simple Harmonic Motion (SHM) lies in its perfect symmetry and the elegant dance between kinetic and potential energy. In this problem, we are asked to find the exact moment when these two energies are perfectly balanced.

Imagine a mass hanging from a spring, oscillating up and down. As it moves through the mean position, its speed is maximum, meaning its kinetic energy is at its peak while the potential energy is zero. Conversely, at the extreme positions, it momentarily stops, meaning its kinetic energy is zero and all the energy is stored as potential energy in the spring.

The question asks for the time when these two energies are exactly equal. Interestingly, the problem provides the mass ($1\text{ kg}$) and the spring constant ($100{\text{ Nm}}^{-1}$). But do we really need them? Let's find out!

We start by writing down the standard expressions for the kinetic energy (KE) and potential energy (PE) of a particle executing SHM at a displacement $y$ from the mean position:

According to the condition given in the problem, we equate the two energies:

Notice how the term $\frac{1}{2}m{\omega }^2$ appears on both sides? We can simply cancel it out! This proves that the specific values of mass and spring constant were just extra information provided to distract us.

We are left with a purely geometric relationship:

This is a profound result. It tells us that the energies are equal not at half the amplitude ($A/2$), but at $1/\sqrt{2}$ (approximately $70.7\%$) of the amplitude.

Now that we know the displacement, we can find the time. The standard equation for displacement in SHM (assuming it starts from the mean position for simplicity, as the time taken to reach $A/\sqrt{2}$ is the same from either the mean or extreme position) is:

Substituting our value for $y$:

We know that the sine function equals $1/\sqrt{2}$ at an angle of $\pi /4$. Therefore:

Recall that the angular frequency $\omega$ is related to the time period $T$ by the equation $\omega =\frac{2\pi }{T}$. Substituting this in:

The problem states that this time is $T/x$. By comparing our result, we can clearly see that:

$$x=8$$

This problem is a classic example of how competitive exams test your conceptual clarity. By giving you the mass and spring constant, the examiners hoped you would waste time calculating the angular frequency ($\omega =\sqrt{k/m}=10\text{ rad/s}$) and the time period ($T=\pi /5\text{ s}$). However, by relying on the fundamental energy equations, we bypassed those calculations entirely and arrived at the elegant, universal truth: in any SHM, the kinetic and potential energies are equal at $T/8$.