Q.The relation between acceleration and displacement of four particles are given below:

The key idea is the defining condition for Simple Harmonic Motion (SHM).

1. For a particle to execute Simple Harmonic Motion, its acceleration must be directly proportional to its displacement from the equilibrium position and always directed towards the equilibrium position.
2. Mathematically, this condition is expressed as $a = -\omega^2 x$, where $a$ is acceleration, $x$ is displacement, and $\omega$ is a positive constant (angular frequency).
3. We examine the given relations:
   • (a) $a_x = +2x$: The acceleration is proportional to $x$, but the positive sign means it's directed away from the equilibrium, not towards it.
   • (b) $a_x = +2x^{2}$: The acceleration is proportional to $x^2$, not $x$.
   • (c) $a_x = -2x^{2}$: The acceleration is proportional to $x^2$, not $x$.
   • (d) $a_x = -2x$: The acceleration is directly proportional to $x$ and the negative sign indicates it is always directed towards the equilibrium position. This matches the form $a = -\omega^2 x$ with $\omega^2 = 2$.

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force acting on the particle is directly proportional to its displacement from the equilibrium position and always directed towards the equilibrium position. This fundamental characteristic is what makes the motion "simple harmonic."

Let's break down what this means for acceleration:

Since force $F = ma$, if $F \propto -x$, then acceleration $a \propto -x$.

The standard mathematical form for the acceleration of a particle undergoing SHM is:

Now, let's examine each given option:

1. Option (a): $a_x = +2x$

   In this case, the acceleration is directly proportional to the displacement, but the sign is positive. This means if the particle is displaced to the right ($x > 0$), the acceleration is also to the right. This would push the particle further away from the equilibrium position, leading to unstable motion, not oscillatory motion. It lacks the restoring nature required for SHM.

2. Option (b): $a_x = +2x^2$

   Here, the acceleration is proportional to the square of the displacement ($x^2$), not the displacement itself ($x$). This is a non-linear relationship. Furthermore, the positive sign indicates it's not a restoring acceleration. Therefore, this is not SHM.

3. Option (c): $a_x = -2x^2$

   In this option, the acceleration is proportional to the square of the displacement ($x^2$). While the negative sign suggests a restoring tendency (acceleration is opposite to the sign of $x^2$, which is always positive, so $a_x$ is always negative), the dependence on $x^2$ makes it non-linear. SHM requires a linear dependence on $x$. This motion might be oscillatory, but it is not simple harmonic motion.

   Watch out

   A common mistake is to only look for the negative sign. While the negative sign is necessary for a restoring force/acceleration, the proportionality must also be linear with displacement ($x$), not $x^2$, $x^3$, or any other power or function of $x$.

4. Option (d): $a_x = -2x$

   This relation perfectly matches the defining equation for SHM, $a_x = -\omega^2 x$.

   • The acceleration is directly proportional to the displacement ($x$).
   • The negative sign indicates that the acceleration is always directed opposite to the displacement, acting as a restoring force towards the equilibrium position. Comparing $a_x = -2x$ with $a_x = -\omega^2 x$, we can see that $\omega^2 = 2$, which means $\omega = \sqrt{2}$ rad/s. This confirms that the particle is executing simple harmonic motion.

Step 1: The defining test for SHM is $a = -\omega^2 x$ — acceleration must be a linear function of displacement with a negative (restoring) sign.

Step 2: Check (a) $a_x=+2x$: linear in $x$, but the sign is positive — this pushes the particle away from equilibrium (unstable), not SHM.

Step 3: Check (b) $a_x=+2x^2$ and (c) $a_x=-2x^2$: both depend on $x^2$, not $x$ — nonlinear, so neither is SHM regardless of sign.

Step 4: Check (d) $a_x=-2x$: linear in $x$ with a negative sign, matching $a=-\omega^2x$ with $\omega^2=2$. This is SHM.

Shortcut: Scan directly for "negative coefficient $\times$ first power of $x$, nothing else" — only (d) has that exact shape.