Animated Solution for Physics - Oscillations: A particle moves with simple harmonic motion in a straight line. In first $\tau$ sec, after starting from rest, it travels a distance $a$ and in next $\tau$ sec, it travels $2a$ in same direction, then

The beauty of Simple Harmonic Motion (SHM) lies in its predictability. If you know where a particle starts and how it moves in the first few moments, you can mathematically decode its entire future and past. This problem is a classic example of using initial conditions and kinematic data to reverse-engineer the core parameters of the motion: the amplitude and the time period.

The problem states that the particle starts from rest. In the realm of SHM, a particle only comes to rest at its extreme positions. Therefore, at time $t=0$, the particle is at an extreme position, which we can denote as $x=A$ (where $A$ is the amplitude).

Because the motion starts from the extreme position, the most natural equation to describe its position $x$ at any time $t$ is the cosine function:

Here, $x$ is the displacement from the mean position, and $\omega$ is the angular frequency.

The problem gives us two critical pieces of information about the distances traveled:

1. In the first $\tau$ seconds: The particle travels a distance $a$. Since it started at $x=A$ and is moving towards the mean position, its new coordinate is $x=A-a$. Substituting this into our SHM equation yields:

2. In the next $\tau$ seconds: The particle travels an additional distance of $2a$ in the same direction. This means the total time elapsed is $2\tau$, and the total distance traveled from the start is $a+2a=3a$. Its new coordinate is $x=A-3a$. Substituting this gives:

We now have a system of two equations with two unknowns ($A$ and $\omega$). The challenge is that the angular terms are $\omega \tau$ and $2\omega \tau$. To bridge this gap, we must call upon a fundamental trigonometric identity—the double angle formula for cosine:

Applying this to our equations, we can express $\cos 2\omega \tau$ in terms of $\cos \omega \tau$:

Now, we roll up our sleeves and execute the algebra. Let's expand the squared term on the right side:

Taking the common denominator on the right side:

We can cancel one $A$ from the denominators (since $A eq 0$) and cross-multiply the remaining $A$ to the left side:

The $A^2$ terms beautifully cancel out, leaving us with a simple linear relationship between $A$ and $a$:

Since the distance $a$ is non-zero, we can divide by $a$ to find the amplitude:

With the amplitude $A$ known, finding the time period $T$ is straightforward. We substitute $A=2a$ back into our first equation:

The principal angle whose cosine is $1/2$ is $\pi /3$ radians. Therefore:

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

Solving for $T$, the $\pi$ terms cancel out, and we get:

This elegant result tells us that the total time period of the oscillation is exactly six times the duration of that first interval. The physics perfectly aligns with the math!