A linear harmonic oscillator of force constant 2 * 10^6 N/m and amplitude 0.01 m has a total mechani

Q: Why isn't the minimum potential energy of the oscillator zero?

The potential energy of a system depends on the chosen reference point. While we often set $U = 0$ at the equilibrium position for simplicity, a system can have a non-zero baseline potential energy. Here, the total energy ($160\text{ J}$) is greater than the oscillation energy ($100\text{ J}$), meaning there is a constant baseline potential energy of $60\text{ J}$ that never converts into kinetic energy.

Q: Does the non-zero minimum potential energy affect the frequency or time period of the SHM?

No, a constant baseline potential energy does not affect the dynamics of the motion. The restoring force is given by $F = -\frac{dU}{dx}$. Since the derivative of a constant baseline is zero, the force and consequently the acceleration, frequency, and time period remain completely unchanged.

Q: How do we know that the maximum kinetic energy is exactly equal to $\frac{1}{2}kA^2$?

Kinetic energy is directly related to the motion (velocity) of the particle. Since the baseline potential energy is constant and cannot be converted into motion, only the dynamic potential energy $\frac{1}{2}kA^2$ is available to be converted into kinetic energy. Thus, $K_{\max} = \frac{1}{2}kA^2$.

Q: What would the energy graphs look like for this system?

Both the kinetic and potential energy graphs are parabolas. The potential energy parabola $U(x)$ has its vertex at $(0, 60\text{ J})$ and opens upwards, reaching $160\text{ J}$ at $x = \pm A$. The kinetic energy parabola $K(x)$ has its vertex at $(0, 100\text{ J})$ and opens downwards, reaching $0\text{ J}$ at $x = \pm A$.

Q: Can the total mechanical energy ever be less than the oscillation energy?

No, the total mechanical energy $E$ must always be greater than or equal to the oscillation energy $E_{\text{osc}}$. If $E < E_{\text{osc}}$, the minimum potential energy $U_{\min}$ would have to be negative, which is mathematically possible but physically means the system has a negative baseline energy relative to the chosen reference.

Question

A linear harmonic oscillator of force constant $2 	imes 10^6 	ext{ N/m}$ and amplitude $0.01 	ext{ m}$ has a total mechanical energy of $160 	ext{ J}$. Its

(a) maximum potential energy is $100 	ext{ J}$

(b) maximum kinetic energy is $100 	ext{ J}$

(c) maximum potential energy is $160 	ext{ J}$

(d) minimum potential energy is zero

Options:

(A) maximum potential energy is $100 	ext{ J}$

(B) maximum kinetic energy is $100 	ext{ J}$

(C) maximum potential energy is $160 	ext{ J}$

(D) minimum potential energy is zero

Sigma Percentile · Accepted Answer

### Introduction to Energy in Simple Harmonic Motion

Simple Harmonic Motion (SHM) is one of the most fundamental and elegant concepts in physics.

When a particle undergoes SHM, energy continuously transitions between kinetic energy ($K$) and potential energy ($U$).

In an ideal, frictionless system, the **Total Mechanical Energy ($E$)** remains constant throughout the motion.

$$
E = K + U = 	ext{constant}
$$

However, a common misconception among students is that the potential energy must always be zero at the mean position.

This problem beautifully highlights how a non-zero reference potential energy affects the overall energy landscape of a harmonic oscillator.

---

### Analyzing the Setup

We are given the following parameters for a linear harmonic oscillator:
- Force constant, $k = 2 	imes 10^6 	ext{ N/m}$
- Amplitude of oscillation, $A = 0.01 	ext{ m}$
- Total mechanical energy, $E = 160 	ext{ J}$

Let's first calculate the energy associated purely with the oscillation itself.

This is the energy required to stretch or compress the spring to its maximum amplitude $A$ relative to the equilibrium position.

$$
E_{	ext{osc}} = \frac{1}{2} k A^2
$$

Substituting the given values:

$$
E_{	ext{osc}} = \frac{1}{2} 	imes (2 	imes 10^6 	ext{ N/m}) 	imes (0.01 	ext{ m})^2
$$

$$
E_{	ext{osc}} = 10^6 	imes 10^{-4} = 100 	ext{ J}
$$

This $100 	ext{ J}$ represents the **dynamic energy** of the oscillation.

It is the maximum kinetic energy the particle can achieve, and it is also the maximum potential energy measured *relative* to the minimum potential energy.

---

### Finding the Maximum Kinetic Energy

At the mean position ($x = 0$), the potential energy of the oscillator is at its minimum value ($U_{\min}$), and the kinetic energy reaches its maximum value ($K_{\max}$).

Since the dynamic oscillation energy is entirely converted into kinetic energy at the mean position:

$$
K_{\max} = E_{	ext{osc}} = 100 	ext{ J}
$$

This directly confirms that **Option (b) is correct**.

---

### Determining the Potential Energy Limits

Now, let's address the potential energy.

The total mechanical energy is the sum of kinetic and potential energy at any point:

$$
E = K(x) + U(x) = 160 	ext{ J}
$$

At the mean position ($x = 0$):
- Kinetic energy is maximum: $K_{\max} = 100 	ext{ J}$
- Potential energy is minimum: $U_{\min}$

Using the conservation of energy:

$$
E = K_{\max} + U_{\min}
$$

$$
160 	ext{ J} = 100 	ext{ J} + U_{\min}
$$

$$
U_{\min} = 60 	ext{ J}
$$

This tells us that the potential energy does not drop to zero at the mean position; instead, it has a baseline value of $60 	ext{ J}$.

Therefore, **Option (d) is incorrect**.

At the extreme positions ($x = \pm A$):
- The particle momentarily comes to rest, so kinetic energy is zero: $K = 0$
- Potential energy reaches its maximum value: $U_{\max}$

Using the conservation of energy again:

$$
E = K_{\min} + U_{\max}
$$

$$
160 	ext{ J} = 0 + U_{\max}
$$

$$
U_{\max} = 160 	ext{ J}
$$

This confirms that **Option (c) is correct**, while Option (a) is incorrect.

---

### Summary of Results

By carefully separating the total mechanical energy into its baseline potential energy and dynamic oscillation energy, we have determined:
1. **Maximum Kinetic Energy** = $100 	ext{ J}$ (Option b)
2. **Maximum Potential Energy** = $160 	ext{ J}$ (Option c)
3. **Minimum Potential Energy** = $60 	ext{ J}$

Thus, the correct options are **(b) and (c)**.

Select Answer:

Visualized Solution

Visualizing the Energy of a Harmonic Oscillator

The Formula for Oscillation Energy

Substituting the Given Values

Calculating the Oscillation Energy

Finding Maximum Kinetic Energy

Analyzing the Minimum Potential Energy

Finding Maximum Potential Energy

The Way Forward

The Sigma Insight: Force and Energy Method in SHM

Introduction to Energy in Simple Harmonic Motion

Analyzing the Setup

Finding the Maximum Kinetic Energy

Determining the Potential Energy Limits

Summary of Results

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