Animated Solution for Physics - Oscillations: A particle is making simple harmonic motion along the X-axis. If at a distances $x_1$ and $x_2$ from the mean position, the velocities of the particle are $v_1$ and $v_2$ respectively, then the time period of its oscillation is given as

Imagine a particle executing Simple Harmonic Motion (SHM) along the X-axis. It dances back and forth around its mean position. We are given two distinct snapshots of this dance: 1. When the particle is at a distance $x_1$ from the mean position, its velocity is $v_1$. 2. When it is at a distance $x_2$, its velocity is $v_2$.

Our mission is to find the time period $T$ of this oscillation using only these two snapshots.

To connect position and velocity, we rely on the fundamental kinematic equation of SHM. The velocity $v$ of a particle at any displacement $x$ is given by:

Here, $A$ is the amplitude of the motion, and $\omega$ is the angular frequency. Since we are dealing with magnitudes and will eventually square the terms, we can drop the $\pm$ sign for convenience.

Applying this master equation to our two specific states, we get a system of two equations:

We have two equations, but we also have two unknowns: the angular frequency $\omega$ and the amplitude $A$. The question asks for the time period (which is directly linked to $\omega$), but it doesn't care about the amplitude $A$. Therefore, our algebraic strategy must be to eliminate $A$.

Square roots make algebraic manipulation cumbersome. Let's clean things up by squaring both sides of our equations:

Now, the path to eliminating $A$ is clear. By subtracting the second equation from the first, the ${\omega }^2{A}^2$ terms will perfectly cancel each other out! Let's perform the subtraction:

Factoring out ${\omega }^2$, we get a beautiful, clean relation:

Isolating ${\omega }^2$, we find:

We are almost at the finish line. We have found ${\omega }^2$, but we need the time period $T$. Recall the fundamental bridge between angular frequency and time period:

Substituting this into our isolated equation, we get:

Taking the square root of both sides yields:

To get $T$ by itself, we simply invert the fraction inside the square root and multiply by $2\pi$:

And there we have it! We have successfully derived the time period using only the given positions and velocities. This technique of squaring and subtracting is a powerful tool in your physics arsenal—keep it sharp!