Animated Solution for Physics - Oscillations: If $x$, $v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T$, then which of the following does not change with time?

Simple Harmonic Motion (SHM) is one of the most beautiful and ubiquitous phenomena in physics. Whether it is a pendulum swinging in a grandfather clock, a mass bouncing on a spring, or the vibrations of atoms in a crystal lattice, the underlying mathematics remains identical.

At any given instant, a particle executing SHM is characterized by three primary kinematic variables: its displacement from the mean position ($x$), its velocity ($v$), and its acceleration ($a$). Because the particle oscillates back and forth, all three of these variables are highly dependent on time, constantly changing as sine or cosine functions.

However, the question challenges us to find a specific combination of these variables that magically strips away the time dependence, leaving behind a pure, unchanging constant.

To unlock this puzzle, we must return to the fundamental defining equation of Simple Harmonic Motion. By definition, a particle is in SHM if its acceleration is directly proportional to its displacement and directed towards the equilibrium position. Mathematically, this is expressed as:

Here, $\omega$ is the angular frequency of the oscillation, which is a constant for a given system. The angular frequency is intimately tied to the time period $T$ of the oscillation through the relation:

Let us systematically evaluate the options provided to see which one yields a constant value.

Let's test option (b), which proposes the expression $\frac{aT}{x}$. We can substitute our master equation $a=-{\omega }^{2}x$ directly into this fraction:

Notice the mathematical elegance here! The time-dependent displacement variable $x$ appears in both the numerator and the denominator. It beautifully cancels out, completely removing the time dependence from the expression:

Now, we substitute $\omega =\frac{2\pi }{T}$ into our simplified expression:

Since the time period $T$ is a strict constant for a given SHM system, the entire expression $-\frac{4{\pi }^2}{T}$ is a constant. It does not change with time. Therefore, option (b) is the correct answer.

For the curious and rigorous student, there is a fascinating historical twist to this specific question from the AIEEE 2009 paper. Let us evaluate option (a): $a^2{T}^2+4{\pi }^2{v}^2$.

Substituting $a=-{\omega }^{2}x$ and $T=\frac{2\pi }{\omega }$, we get:

Factoring out $4{\pi }^2$, we get:

From the energy conservation principle in SHM, we know the velocity-displacement relation is $v^2={\omega }^2(A^2-x^2)$, which rearranges to $v^2+{\omega }^2{x}^2={\omega }^2{A}^2$. Substituting this back:

Astoundingly, option (a) is ALSO a constant! This was a famous error in the examination where two options were technically correct. However, option (b) is the most direct, algebraically simple answer and is the one officially recognized in standard solution keys.