For a particle executing SHM the displacement x is given by x = A cos(ωt). Identify the graph which

Q: Why is the potential energy graph versus displacement a parabola?

Because the potential energy formula is $PE = \frac{1}{2} k x^2$, which is of the quadratic form $y = c x^2$. This mathematically represents a parabola opening upwards with its vertex at the origin $(0,0)$.

Q: Why does the PE vs $t$ graph start at a maximum value?

At $t = 0$, the displacement is $x = A \cos(0) = A$, which is the extreme position. At the extreme position, the velocity is zero (minimum kinetic energy) and the displacement is maximum, meaning all mechanical energy is stored as potential energy. Thus, PE is maximum at $t = 0$.

Q: What is the frequency of oscillation of the potential energy compared to the displacement?

The potential energy oscillates at twice the frequency of the displacement. Since $PE \propto \cos^2 \omega t = \frac{1 + \cos 2\omega t}{2}$, the angular frequency of the potential energy is $2\omega$, which is double the frequency of the displacement oscillation.

Q: Why can't potential energy be negative in this problem?

Potential energy is proportional to $x^2$. Since any real number squared is non-negative, and the spring constant $k$ is positive, the potential energy $PE = \frac{1}{2} k x^2$ must always be greater than or equal to zero.

Q: How would the kinetic energy graphs look?

Kinetic energy is given by $KE = \frac{1}{2} k (A^2 - x^2)$ as a function of displacement, which is an inverted parabola (Curve IV). As a function of time, $KE = \frac{1}{2} k A^2 \sin^2 \omega t$, which starts at zero at $t = 0$ (Curve II).

Question

For a particle executing SHM the displacement $x$ is given by $x = A \cos \omega t$. Identify the graph which represents the variation of potential energy (PE) as a function of time $t$ and displacement $x$.

Options:

(A) I, III

(B) II, IV

(C) II, III

(D) I, IV

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Simple Harmonic Motion (SHM) is one of the most elegant and symmetric phenomena in classical mechanics.

In this problem, we are given a particle executing SHM whose displacement is mathematically described by the equation:

$$x = A \cos \omega t$$

This equation tells us that at $t = 0$, the particle is located at its positive extreme position, $x = A$.

As time progresses, the particle oscillates back and forth between $x = +A$ and $x = -A$.

Our goal is to identify the correct graphs representing the potential energy (PE) of this system as a function of both time $t$ and displacement $x$.

### The Physics of Potential Energy

To understand how potential energy behaves, we must look at its fundamental definition for a simple harmonic oscillator.

The restoring force acting on the particle is conservative and is given by Hooke's Law, $F = -kx$.

The potential energy associated with this force is:

$$PE = \frac{1}{2} k x^2$$

Here, $k = m \omega^2$ is the force constant of the oscillator.

Looking at this equation, we can immediately make two crucial observations:

1. **Non-negativity**: Since $x$ is squared, the potential energy can never be negative. It is always greater than or equal to zero.

2. **Parabolic Shape**: The graph of potential energy versus displacement $x$ is a quadratic function of the form $y = c x^2$. This represents a parabola opening upwards with its vertex at the origin $(0,0)$.

Let's evaluate the potential energy at key positions:

- At the mean position ($x = 0$):

$$PE = \frac{1}{2} k (0)^2 = 0$$

- At the extreme positions ($x = \pm A$):

$$PE = \frac{1}{2} k (\pm A)^2 = \frac{1}{2} k A^2$$

This tells us that the potential energy is zero at the center and reaches its maximum value at the boundaries.

Comparing this with the given displacement graphs, **Curve III** perfectly represents this parabolic behavior, starting at $0$ at $x = 0$ and rising symmetrically to a maximum at $x = \pm A$.

### Time-Dependent Behavior

Now, let's analyze how the potential energy changes with time.

By substituting the displacement equation $x = A \cos \omega t$ into our potential energy formula, we get:

$$PE(t) = \frac{1}{2} k (A \cos \omega t)^2 = \frac{1}{2} k A^2 \cos^2 \omega t$$

To find the starting point of this graph, let's evaluate it at $t = 0$:

$$PE(0) = \frac{1}{2} k A^2 \cos^2(0) = \frac{1}{2} k A^2$$

Since $\cos(0) = 1$, the potential energy at $t = 0$ is at its maximum value.

This makes physical sense because at $t = 0$, the particle is at the extreme position $x = A$, where it momentarily stops, meaning its kinetic energy is zero and all its mechanical energy is stored as potential energy.

Looking at the time-dependent graphs:

- **Curve I** starts at a maximum value at $t = 0$.

- **Curve II** starts at zero at $t = 0$.

Therefore, **Curve I** is the correct representation of potential energy as a function of time.

### The Final Verdict

By combining our two analyses, we conclude:

- The potential energy versus time $t$ is represented by **Curve I**.

- The potential energy versus displacement $x$ is represented by **Curve III**.

Thus, the correct pair of graphs is **I and III**, which corresponds to **Option (a)**.

Animated Solution for Physics - Oscillations: For a particle executing SHM the displacement $x$ is given by $x = A cos ω t$ . Identify the graph which represents the variation of potential energy (PE) as a function of time $t$ and displacement $x$ .

Select Answer:

Visualized Solution

Understanding the Physical System

The Formula for Potential Energy

Analyzing PE as a Function of Displacement $x$

Analyzing PE as a Function of Time $t$

Evaluating PE at $t = 0$

Identifying the Correct Option

Extension: Kinetic Energy Graphs

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Physics of Potential Energy

Time-Dependent Behavior

The Final Verdict

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