Energy in simple harmonic motion

1. Overview

In an ideal simple harmonic motion (SHM) system, the total energy ($E_{total}$) remains constant over time, provided there are no dissipative forces such as friction or air resistance. This principle is a direct application of the Law of Conservation of Energy. Within the system, energy is continuously and periodically transformed between kinetic energy ($E_k$) and potential energy ($E_p$).

The maximum value of kinetic energy occurs at the equilibrium position, where the velocity is at its peak. Conversely, the maximum value of potential energy occurs at the maximum displacement (amplitude), where the oscillator momentarily comes to rest. At any point in the oscillation, the sum of these two energies equals the total energy of the system: $E_{total} = E_k + E_p$.

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Key Definitions

• Simple Harmonic Motion (SHM): A periodic motion where the acceleration of the object is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.
• Restoring Force: The resultant force that acts on an oscillator to return it to the equilibrium position. In SHM, $F = -kx$ (where $k = m\omega^2$).
• Amplitude ($x_0$): The maximum displacement of the oscillator from its equilibrium position.
• Angular Frequency ($\omega$): The rate of change of angular displacement, measured in radians per second ($rad\ s^{-1}$). It is defined as $\omega = 2\pi f = \frac{2\pi}{T}$.
• Equilibrium Position: The central point of the motion ($x = 0$) where the resultant force on the object is zero and its velocity is maximum.
• Potential Energy ($E_p$): The energy stored in the system due to the position or configuration of the oscillator (e.g., elastic potential energy in a spring or gravitational potential energy in a pendulum).
• Kinetic Energy ($E_k$): The energy possessed by the oscillator due to its motion.

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Content

3.1 The Energy Interchange Process

The interchange between kinetic and potential energy is the defining characteristic of an oscillating system.

1. At the Equilibrium Position ($x = 0$):

   • The displacement is zero, meaning the restoring force is zero. Consequently, the potential energy is zero ($E_p = 0$).
   • The oscillator is moving at its maximum velocity ($v_{max} = \omega x_0$).
   • The kinetic energy is at its maximum ($E_k = E_{total}$).

2. At Maximum Displacement/Amplitude ($x = \pm x_0$):

   • The oscillator reaches its furthest point and momentarily stops to change direction ($v = 0$).
   • The kinetic energy is zero ($E_k = 0$).
   • The restoring force and displacement are at their maximum, meaning the potential energy is at its maximum ($E_p = E_{total}$).

3. At Intermediate Positions ($0 < |x| < x_0$):

   • The system possesses a mix of both kinetic and potential energy.
   • As the object moves from the amplitude toward equilibrium, $E_p$ is converted into $E_k$.
   • As the object moves from equilibrium toward the amplitude, $E_k$ is converted into $E_p$.

3.2 Mathematical Derivations

To solve exam problems, you must understand how the energy equations are derived from the fundamental SHM kinematic equations.

1. Kinetic Energy ($E_k$): The velocity $v$ of an object in SHM at any displacement $x$ is given by the formula (found on the data sheet): $$v = \pm \omega \sqrt{x_0^2 - x^2}$$

The standard formula for kinetic energy is: $$E_k = \frac{1}{2} m v^2$$

Substituting the SHM velocity expression into the kinetic energy formula: $$E_k = \frac{1}{2} m (\omega \sqrt{x_0^2 - x^2})^2$$ $$E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)$$

2. Total Energy ($E_{total}$): The total energy is constant and is equal to the kinetic energy when the object is at the equilibrium position ($x = 0$). $$E_{total} = \frac{1}{2} m \omega^2 (x_0^2 - 0^2)$$ $$E_{total} = \frac{1}{2} m \omega^2 x_0^2$$ Note: This formula shows that the total energy of an undamped oscillator is proportional to the mass, the square of the angular frequency, and the square of the amplitude.

3. Potential Energy ($E_p$): Using the principle of conservation of energy: $E_{total} = E_k + E_p$. Therefore, $E_p = E_{total} - E_k$. $$E_p = \frac{1}{2} m \omega^2 x_0^2 - \left[ \frac{1}{2} m \omega^2 (x_0^2 - x^2) \right]$$ $$E_p = \frac{1}{2} m \omega^2 x_0^2 - \frac{1}{2} m \omega^2 x_0^2 + \frac{1}{2} m \omega^2 x^2$$ $$E_p = \frac{1}{2} m \omega^2 x^2$$

3.3 Energy-Displacement Graphs

The relationship between energy and displacement is quadratic, resulting in parabolic curves.

• Potential Energy ($E_p$): An upward-opening parabola with its minimum ($0$) at $x = 0$ and its maximum at $x = \pm x_0$.
• Kinetic Energy ($E_k$): A downward-opening (inverted) parabola with its maximum at $x = 0$ and its minimum ($0$) at $x = \pm x_0$.
• Total Energy ($E_{total}$): A horizontal straight line, representing the constant sum of $E_k$ and $E_p$.

Intersection Points: The $E_k$ and $E_p$ curves intersect when $E_k = E_p$. This occurs when: $$\frac{1}{2} m \omega^2 (x_0^2 - x^2) = \frac{1}{2} m \omega^2 x^2$$ $$x_0^2 - x^2 = x^2 \implies 2x^2 = x_0^2 \implies x = \pm \frac{x_0}{\sqrt{2}} \approx \pm 0.707 x_0$$ At this displacement, the energy is split exactly $50/50$ between kinetic and potential.

3.4 Energy-Time Relationships

If the displacement follows a sinusoidal pattern (e.g., $x = x_0 \sin(\omega t)$), the energy components also vary sinusoidally but with a squared relationship.

• $E_p = \frac{1}{2} m \omega^2 x_0^2 \sin^2(\omega t)$
• $E_k = \frac{1}{2} m \omega^2 x_0^2 \cos^2(\omega t)$

The Double Frequency Rule: In one complete oscillation of the object (period $T$), the kinetic energy reaches its maximum twice (once moving "forward" through equilibrium and once moving "backward"). Consequently:

• The period of the energy oscillation is $\frac{T}{2}$.
• The frequency of the energy oscillation is $2f$.
• Energy is always a positive quantity; the graphs never drop below the horizontal axis.

3.5 Worked Example 1 — Calculating Energy from System Parameters

Question: A horizontal mass-spring system consists of a $250\ g$ mass attached to a spring. The system undergoes SHM with a frequency of $2.5\ Hz$ and an amplitude of $4.0\ cm$.

1. Calculate the total energy of the system.
2. Calculate the potential energy when the mass is $2.0\ cm$ from the equilibrium position.

Step 1: Convert units to SI.

• $m = 0.250\ kg$
• $f = 2.5\ Hz$
• $x_0 = 0.040\ m$
• $x = 0.020\ m$

Step 2: Calculate angular frequency $\omega$. $$\omega = 2\pi f = 2 \times \pi \times 2.5 = 5\pi \approx 15.71\ rad\ s^{-1}$$

Step 3: Calculate Total Energy ($E_{total}$). $$E_{total} = \frac{1}{2} m \omega^2 x_0^2$$ $$E_{total} = 0.5 \times 0.250 \times (15.71)^2 \times (0.040)^2$$ $$E_{total} = 0.125 \times 246.74 \times 0.0016$$ $$E_{total} = 0.04935\ J \approx 0.049\ J\ (2\ s.f.)$$

Step 4: Calculate Potential Energy ($E_p$) at $x = 0.020\ m$. $$E_p = \frac{1}{2} m \omega^2 x^2$$ $$E_p = 0.5 \times 0.250 \times (15.71)^2 \times (0.020)^2$$ $$E_p = 0.125 \times 246.74 \times 0.0004$$ $$E_p = 0.01234\ J \approx 0.012\ J\ (2\ s.f.)$$

3.6 Worked Example 2 — Finding Amplitude from Energy

Question: A simple pendulum of mass $500\ g$ has a total energy of $0.15\ J$. If the angular frequency of the oscillation is $4.2\ rad\ s^{-1}$, determine the amplitude of the oscillation.

Step 1: Identify known values.

• $E_{total} = 0.15\ J$
• $m = 0.500\ kg$
• $\omega = 4.2\ rad\ s^{-1}$

Step 2: Rearrange the total energy formula for $x_0$. $$E_{total} = \frac{1}{2} m \omega^2 x_0^2 \implies x_0^2 = \frac{2 E_{total}}{m \omega^2}$$ $$x_0 = \sqrt{\frac{2 E_{total}}{m \omega^2}}$$

Step 3: Substitute and solve. $$x_0 = \sqrt{\frac{2 \times 0.15}{0.500 \times (4.2)^2}}$$ $$x_0 = \sqrt{\frac{0.30}{0.500 \times 17.64}}$$ $$x_0 = \sqrt{\frac{0.30}{8.82}}$$ $$x_0 = \sqrt{0.03401} \approx 0.1844\ m$$ Answer: $x_0 = 0.18\ m$ or $18\ cm$ (2 s.f.)

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Key Equations

Equation	Description	Status
$E_{total} = \frac{1}{2} m \omega^2 x_0^2$	Total energy of the SHM system.	Must Memorize
$E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)$	Kinetic energy at displacement $x$.	Must Memorize
$E_p = \frac{1}{2} m \omega^2 x^2$	Potential energy at displacement $x$.	Must Memorize
$v = \pm \omega \sqrt{x_0^2 - x^2}$	Velocity at displacement $x$.	Data Sheet
$\omega = 2\pi f = \frac{2\pi}{T}$	Relationship between $\omega, f,$ and $T$.	Data Sheet

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Common Mistakes to Avoid

• ❌ Wrong: Confusing $x$ (instantaneous displacement) with $x_0$ (amplitude) in the total energy formula.
  • ✓ Right: $E_{total}$ always uses the amplitude ($x_0$). If you use a smaller $x$, you are only calculating the potential energy at that specific point.
• ❌ Wrong: Forgetting to square the angular frequency ($\omega$) or the displacement/amplitude.
  • ✓ Right: Always perform a unit check. Energy is $kg \cdot (rad\ s^{-1})^2 \cdot m^2$, which simplifies to $kg\ m^2\ s^{-2}$ (Joules).
• ❌ Wrong: Using the frequency $f$ in $Hz$ directly in the energy equation.
  • ✓ Right: The formula requires angular frequency $\omega$. You must multiply $f$ by $2\pi$ first.
• ❌ Wrong: Drawing energy-displacement graphs with straight lines (V-shapes).
  • ✓ Right: These graphs must be curves (parabolas) because the energy depends on the square of the displacement.
• ❌ Wrong: Assuming the frequency of energy change is the same as the frequency of the motion.
  • ✓ Right: The energy reaches a maximum twice per cycle, so the energy frequency is $2f$.

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Exam Tips

1. Conservation of Energy: If an exam question mentions "no resistive forces" or "undamped," immediately state that the total energy remains constant. This is often the first mark in a multi-part calculation.
2. Graph Interpretation: You may be asked to identify curves on an energy-displacement graph. Remember:
   • The curve that is zero at the center is Potential Energy ($E_p$).
   • The curve that is maximum at the center is Kinetic Energy ($E_k$).
3. The $x_0/\sqrt{2}$ Point: Be aware that $E_k = E_p$ does not happen at half the amplitude ($x_0/2$). It happens at approximately $0.71x_0$. At $x = 0.5x_0$, the potential energy is only $25%$ of the total energy ($E_p = \frac{1}{2} m \omega^2 (0.5x_0)^2 = 0.25 E_{total}$).
4. Unit Consistency: Cambridge 9702 papers often provide mass in grams ($g$) and amplitude in centimeters ($cm$) or millimeters ($mm$). Always convert to $kg$ and $m$ before starting your calculation to ensure your answer is in Joules ($J$).
5. Derivation Skills: If you forget the total energy formula, remember it is just the maximum kinetic energy. Use $v_{max} = \omega x_0$ and plug it into $E_k = \frac{1}{2}mv^2$. This derivation is frequently required in structured questions.