Animated Solution for Physics - Oscillations: A block of mass $5\text{ kg}$ moves along the x-direction subject to the force $F=(-20x+10)\text{ N}$, with the value of $x$ in metre. At time $t=0\text{ s}$, it is at rest at position $x=1\text{ m}$. The position and momentum of the block at $t=\left(\frac{\pi }{4}\right)\text{ s}$ are

The beauty of physics lies in its ability to predict the future—at least, the future of a block moving under a specific force! In this problem, we are given a block of mass $5\text{ kg}$ subjected to a position-dependent force $F=-20x+10$. Our goal is to determine its exact position and momentum at a specific time $t=\frac{\pi }{4}\text{ s}$. Let's embark on this kinematic journey.

The very first clue lies in the nature of the force. The force $F$ depends linearly on the position $x$. In mechanics, whenever you see a force that tries to pull an object back towards a central point, your "Simple Harmonic Motion" (SHM) radar should start beeping.

To confirm this, we need to look at the acceleration. According to Newton's Second Law, acceleration is force divided by mass:

Simplifying this expression, we get:

To reveal the true nature of this motion, let's factor out the $-4$ from our acceleration equation:

This is a beautiful revelation! The standard equation for Simple Harmonic Motion is $a=-{\omega }^2(x-x_0)$, where $\omega$ is the angular frequency and $x_0$ is the mean (equilibrium) position. By comparing our equation to the standard form, we can immediately extract two vital pieces of information:

1. Angular Frequency: ${\omega }^2=4⟹\omega =2\text{ rad/s}$. 2. Mean Position: $x_0=0.5\text{ m}$.

This means the block is oscillating back and forth around the point $x=0.5\text{ m}$.

Now that we know it's SHM, we can write the general equation for its position as a function of time:

We need to find the amplitude $A$ and the initial phase $\varphi$. The problem states that at $t=0$, the block is at rest ($v=0$) at position $x=1\text{ m}$. In SHM, the points where the velocity is zero are the extreme positions.

Since the block is released from rest at $x=1\text{ m}$, this is our positive extreme. The amplitude is simply the distance from the mean position to this extreme:

Because it starts exactly at the positive extreme, the cosine function perfectly models this without any phase shift, meaning $\varphi =0$. Plugging everything into our general equation, we get the master position equation:

We are asked to find the position and momentum at $t=\frac{\pi }{4}\text{ s}$. Let's start with the position. Substituting the time into our equation:

Since $\cos \left(\frac{\pi }{2}\right)=0$, the position is exactly $0.5\text{ m}$. The block is passing right through its mean position!

Next, we need the momentum, which requires the velocity. We find velocity by differentiating the position equation with respect to time:

Evaluating this at $t=\frac{\pi }{4}\text{ s}$:

The negative sign tells us the block is moving in the negative x-direction. Finally, momentum $P$ is mass times velocity:

Conclusion: At $t=\frac{\pi }{4}\text{ s}$, the block is at $x=0.5\text{ m}$ with a momentum of $-5\text{ kg m/s}$. This perfectly matches option (C).