Animated Solution for Physics - Oscillations: A particle executes simple harmonic motion between $x=-A$ and $x=+A$. The time taken for it to go from $O$ to $A/2$ is $T_1$ and to go from $A/2$ to $A$ is $T_2$, then

Simple Harmonic Motion (SHM) is one of the most fundamental and beautiful concepts in physics. From the gentle sway of a grandfather clock's pendulum to the rapid vibration of atoms in a crystal lattice, SHM governs periodic motion across all scales of the universe.

At its heart, SHM is characterized by a restoring force that is directly proportional to the displacement from a central equilibrium position. This simple linear relationship leads to a rich tapestry of mathematical and physical consequences.

In this article, we will explore a classic problem that tests our deep understanding of time and velocity in SHM. We are asked to compare the time taken by a particle to cover two equal spatial intervals: from the mean position $O$ to half the amplitude $A/2$, and from half the amplitude $A/2$ to the extreme position $A$.

Let us visualize a particle executing simple harmonic motion along a straight line. The center of this motion is the mean position, which we designate as $O$ ($x=0$). The particle oscillates symmetrically between the extreme positions $x=-A$ and $x=+A$.

We are interested in two distinct intervals of motion:

1. Interval 1: Moving from $x=0$ to $x=A/2$. The time taken for this is $T_1$. 2. Interval 2: Moving from $x=A/2$ to $x=A$. The time taken for this is $T_2$.

At first glance, a student might be tempted to think that since both intervals cover the exact same distance of $A/2$, the time taken should be equal ($T_1=T_2$). However, this assumption relies on the particle moving at a constant speed, which is not the case in simple harmonic motion.

Velocity Profile

Before diving into the rigorous mathematics, let us build a strong physical intuition. How does the velocity of a particle in SHM vary as it moves from the mean position to the extreme position?

The velocity $v$ of a particle at any position $x$ is given by the well-known relation:

where $\omega$ is the angular frequency of oscillation.

Let us analyze this equation at key points:

At the mean position ($x=0$), the velocity is at its maximum value:

$$v_{max}=\omega A$$

At the extreme position ($x=A$), the velocity drops to zero:

As the particle travels from $O$ to $A$, it is continuously decelerating.

In the first half of the journey (from $0$ to $A/2$), the particle is moving at its highest speeds. In the second half of the journey (from $A/2$ to $A$), the particle slows down significantly as it approaches the turning point.

Since the average speed in the first interval is much higher than the average speed in the second interval, it must take less time to cover the first interval. Therefore, we can intuitively conclude that:

This is a powerful qualitative way to solve the problem in seconds during an exam!

Now, let us back up our intuition with precise mathematical calculations. Since the particle starts from the mean position $O$ at $t=0$, we can represent its displacement $x(t)$ using a sine function:

Let us calculate $T_1$, the time taken to reach $x=A/2$:

Dividing both sides by $A$:

Since the particle is moving from the mean position to the extreme for the first time, we take the principal value:

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

Thus, the particle spends exactly one-twelfth of its total time period traveling from $O$ to $A/2$.

Next, let us find the total time taken to reach the extreme position $x=A$. This total time is $T_1+T_2$:

This makes perfect sense! It takes exactly one-quarter of the total time period to complete one-fourth of a full oscillation cycle (from mean to extreme).

Now, we can easily solve for $T_2$ by subtraction:

To subtract these fractions, we find a common denominator:

Comparing our two time intervals:

$T_1=\frac{T}{12}$ $T_2=\frac{T}{6}$

Since $\frac{T}{12}<\frac{T}{6}$, we have rigorously proven that:

In fact, we can see that $T_2$ is exactly twice as large as $T_1$ ($T_2=2T_1$).

This problem teaches us a vital lesson about non-uniform motion. In SHM, equal spatial intervals do not correspond to equal time intervals.

Always remember the standard time divisions for a particle starting from the mean position: $0\to \frac{A}{2}$ takes $\frac{T}{12}$ $\frac{A}{2}\to A$ takes $\frac{T}{6}$ * $0\to A$ takes $\frac{T}{4}$

Mastering these standard intervals will allow you to visualize and solve complex SHM problems with incredible speed and confidence!