Animated Solution for Physics - Laws of Motion: A particle of mass $0.3\text{ kg}$ is subjected to a force $F=-kx$ with $k=15{\text{ Nm}}^{-1}$. What will be its initial acceleration, if it is released from a point $20\text{ cm}$ away from the origin?

A Particle on a Spring

Imagine a smooth, frictionless horizontal surface. On this surface, we have a small particle with a mass of $0.3\text{ kg}$. This particle is attached to one end of a spring, while the other end of the spring is firmly fixed to a wall at the origin, where $x=0$.

Now, we take this particle and pull it away from the origin, stretching the spring until the particle reaches a position of $x=20\text{ cm}$. At this exact moment, we are holding it still. The velocity is zero, but the spring is stretched and eager to snap back.

What happens the very instant we let go? That is the core of our problem. We need to find the initial acceleration of the particle right at the moment of release.

Hooke's Law

When you stretch a spring, you can feel it pulling back against your hand. This is known as the restoring force. In physics, for an ideal spring, this force is described by Hooke's Law.

Hooke's Law states that the restoring force exerted by a spring is directly proportional to its displacement from the equilibrium position. Mathematically, it is written as:

Here, $k$ is the spring constant, which tells us how stiff the spring is. In our case, $k=15\text{ N/m}$. The variable $x$ is the displacement from the origin.

But what about that negative sign? The minus sign is crucial! It indicates that the force always acts in the opposite direction to the displacement. If you pull the particle to the right (positive $x$), the spring pulls it back to the left (negative force). This is why it's called a restoring force—it always wants to restore the system to equilibrium.

Newton's Second Law

Now we know the force acting on the particle. But how does force relate to motion? For that, we turn to the legendary Sir Isaac Newton.

Newton's Second Law of Motion tells us that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Since the restoring force of the spring is the only horizontal force acting on our particle (ignoring any air resistance or friction), we can equate these two expressions for force.

By bringing Hooke's Law and Newton's Second Law together, we get our master equation for this system:

We want to find the initial acceleration, $a$. So, let's isolate $a$ by dividing both sides by the mass, $m$:

This beautiful little equation tells us exactly how the acceleration depends on the position of the particle. Notice that the acceleration is not constant! As the particle moves and $x$ changes, the acceleration $a$ will also change. However, we are only interested in the initial acceleration, right at the moment of release when $x=20\text{ cm}$.

Before we rush into plugging in the numbers, we must be very careful. Physics problems love to set traps with units!

Let's look at our given values: - Mass, $m=0.3\text{ kg}$ (Standard SI unit, good!) - Spring constant, $k=15\text{ N/m}$ (Standard SI unit, good!) - Displacement, $x=20\text{ cm}$ (Wait a minute!)

The displacement is given in centimeters, but our spring constant is in Newtons per meter. If we multiply these directly, our units will clash, and our answer will be completely wrong. We must convert the displacement into meters.

Since there are $100\text{ cm}$ in a meter, we divide by $100$:

Now that our units are perfectly aligned, we can confidently substitute the values into our acceleration equation:

Let's compute the numerator first. Multiplying $15$ by $0.2$ is the same as finding one-fifth of $15$, which is $3$.

Finally, dividing $3$ by $0.3$ gives us $10$.

We have arrived at our answer: $a=-10{\text{ m/s}}^2$.

But what does this mean physically? The magnitude of the acceleration is $10{\text{ m/s}}^2$. This is the value we were looking for, matching option (d).

The negative sign is simply a directional indicator. It tells us that while the particle is located at a positive position ($+20\text{ cm}$), the acceleration vector is pointing in the negative direction (towards the left, back to the origin).

As the particle flies towards the origin, the value of $x$ will decrease, meaning the force and acceleration will also decrease. By the time it crosses the origin ($x=0$), the acceleration will be exactly zero, but its velocity will be at its maximum! This beautiful, continuous exchange between force, acceleration, and velocity is the very heartbeat of simple harmonic motion.