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Energy of Simple Harmonic Oscillators

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, this topic is about how energy in an oscillating system, like a bouncing spring, swaps back and forth between motion energy (kinetic) and stored energy (potential), while the total amount of energy stays constant.

Why this matters

Think about the last time you were on a swing set at a park. You get a push and start swinging back and forth. Picture that moment when you're at the very peak of your swing—you feel weightless for a split second, completely still before you start falling forward again. At that peak, you're not moving, but you have a lot of stored energy because of your height.

Now, think about the bottom of the arc. Whoosh! That's where you're moving the fastest. All that stored energy from the peak has turned into the energy of motion. You trade height for speed, and then speed for height, over and over.

That constant trade-off is the heart of what we're studying today: the energy of a simple harmonic oscillator. We'll see how this energy swapping works, how to calculate it, and how it always, always adds up to the same total.

Diagram

Energy of a Simple Harmonic Oscillator A graph showing the relationship between position and energy for a simple harmonic oscillator. The x-axis represents position from -A to +A. The y-axis represents energy. A flat horizontal line shows the constant total energy (E). A U-shaped parabola shows the potential energy (U), which is zero at the center and maximum at the ends. An inverted parabola shows the kinetic energy (K), which is maximum at the center and zero at the ends. At any position, the sum of K and U equals E. Position (x) Energy (J) -A 0 +A E U s K U K x
This diagram shows a graph of energy versus position for a simple harmonic oscillator. The horizontal axis is position, and the vertical axis is energy. A flat, dashed line represents the constant total energy. A yellow, U-shaped parabola represents potential energy, while a green, inverted parabola represents kinetic energy. At any position, the sum of the kinetic and potential energies equals the total energy.

Concept map

flowchart TD
    A[At Amplitude x = +A] --> B(v = 0 --> K = 0<br/>x = max --> U = max);
    B --> C{Moving toward equilibrium};
    C --> D(v increases --> K increases<br/>x decreases --> U decreases);
    D --> E[At Equilibrium x = 0];
    E --> F(v = max --> K = max<br/>x = 0 --> U = 0);
    F --> G{Moving toward other amplitude x = -A};
    G --> H(v decreases --> K decreases<br/>x increases --> U increases);
    H --> I[At Amplitude x = -A];
    I --> J(v = 0 --> K = 0<br/>x = max --> U = max);
    J --> C;

Core explanation

Hello everyone! Today we're diving into one of my favorite applications of a concept you already know and love: conservation of energy. We're just applying it to our new context of simple harmonic motion (SHM).

The Total Energy is Constant

Remember from our unit on energy that for a closed system with no friction or air resistance, the total mechanical energy E is always conserved. It's the sum of the kinetic energy K and the potential energy U.

E_total = K + U

For an object oscillating in SHM, like a block on a frictionless horizontal spring, this principle holds true. The total energy of the system remains constant throughout the entire motion. The energy just changes its form.

Think of it like this: you have a $20 gift card for a coffee shop. You can spend it on a latte (kinetic energy) or a muffin (potential energy). You might buy just a latte one day, just a muffin another, or a small coffee and a mini-scone on a third day. No matter what you buy, the total value you can get is always capped at $20. The total energy in an oscillator works the same way; it's a fixed budget that gets spent on K and U.

The Two Forms of Energy in SHM

Let's get specific for our classic example: a mass m on a horizontal spring with spring constant k.

  1. 1
    Kinetic Energy (K)
    This is the energy of motion. You know this one well. K = (1/2)mv² The block moves fastest when it passes through the equilibrium position (x = 0). So, kinetic energy is maximum at x = 0. At the endpoints of the motion (the amplitudes, x = +A and x = -A), the block momentarily stops to turn around. Its velocity is zero, so the kinetic energy is zero at x = ±A.
  2. 2
    Elastic Potential Energy (U)
    This is the energy stored in the spring when it's stretched or compressed. U_s = (1/2)kx² The spring is most stretched or compressed at the amplitudes (x = ±A). So, potential energy is maximum at x = ±A. At the equilibrium position (x = 0), the spring is in its natural, relaxed state, so the potential energy is zero at x = 0.

Putting It All Together: The Energy Trade-Off

Let's trace the energy through one full oscillation.

  • At the Amplitude (x = A): The mass is momentarily at rest (v = 0), so K = 0. All the system's energy is stored in the stretched spring. E_total = K + U = 0 + (1/2)kA² So, the total energy of the system is determined by the spring constant and the square of the amplitude: E_total = (1/2)kA²

    This is a crucial equation. It tells you the total energy of the oscillator at any point, because it's constant!

  • At the Equilibrium Position (x = 0): The spring is not stretched (x = 0), so U = 0. All the system's energy has been converted into kinetic energy, and the mass is moving at its maximum speed, v_max. E_total = K + U = (1/2)mv_max² + 0 Since E_total is constant, we can set the energy at the amplitude equal to the energy at equilibrium: (1/2)kA² = (1/2)mv_max²

  • At any other position x: The energy is a mix of both kinetic and potential. E_total = (1/2)mv² + (1/2)kx² And since we know E_total is always equal to (1/2)kA², we can write: (1/2)kA² = (1/2)mv² + (1/2)kx²

This single equation is incredibly powerful. If you know k, A, and m, you can find the speed v of the block at any position x.

Visualizing Energy with a Graph

This is where it all clicks into place. If we plot the energy of the system as a function of position x, we get a beautiful graph.

Energy vs. Position Graph

  • The Potential Energy U ((1/2)kx²) is a parabola that opens upward, hitting zero at x=0 and reaching its maximum value at x=±A.
  • The Total Energy E ((1/2)kA²) is a constant, so it's a flat horizontal line.
  • The Kinetic Energy K must be the difference between the total and the potential (K = E - U). This makes it an upside-down parabola. It's maximum at x=0 (where U is zero) and zero at x=±A (where U equals E).

Worked examples

Let's walk through a couple of problems to make sure this is crystal clear.

Example 1

Calculating Energy and Speed

Problem: A 0.50 kg block is attached to a horizontal spring with a spring constant of 80 N/m. The block is pulled to a position of x = 0.10 m from equilibrium and released from rest. (a) What is the total mechanical energy of the system? (b) What is the maximum speed of the block? (c) What is the speed of the block when it is at position x = 0.05 m?

Solution:

(a) Finding Total Energy

  • What do we know?
    The block is released from rest at x = 0.10 m. This position is the amplitude, A, because the initial velocity is zero. We have m = 0.50 kg, k = 80 N/m, and A = 0.10 m.
  • Why this step?
    The easiest place to calculate total energy is at an amplitude, where kinetic energy is zero. The total energy is equal to the maximum potential energy.
  • Calculation
    E_total = (1/2)kA² E_total = (1/2)(80 N/m)(0.10 m)² E_total = (1/2)(80)(0.01) = 0.40 J The total energy of the system is 0.40 Joules. This value is constant throughout the oscillation.

(b) Finding Maximum Speed

  • Why this step?
    The maximum speed (v_max) occurs at the equilibrium position (x = 0), where all the energy is kinetic. We can set the total energy we just found equal to the maximum kinetic energy.
  • Calculation
    E_total = K_max = (1/2)mv_max² 0.40 J = (1/2)(0.50 kg)v_max² 0.40 = (0.25)v_max² v_max² = 0.40 / 0.25 = 1.6 v_max = √1.6 ≈ 1.26 m/s

(c) Finding Speed at a Specific Position

  • Why this step?
    At any position that isn't an endpoint or equilibrium, the total energy is a mix of kinetic and potential. We'll use the full conservation of energy equation.
  • Calculation
    E_total = K + U = (1/2)mv² + (1/2)kx² We know E_total = 0.40 J, so: 0.40 J = (1/2)(0.50 kg)v² + (1/2)(80 N/m)(0.05 m)² 0.40 = (0.25)v² + (1/2)(80)(0.0025) 0.40 = (0.25)v² + 0.10 0.30 = (0.25)v² v² = 0.30 / 0.25 = 1.2 v = √1.2 ≈ 1.10 m/s
  • Common Mistake Alert
    A student might try to use kinematics here (v = v₀ + at). Don't do it! The acceleration is not constant in SHM, so those equations don't apply. Always, always use energy for this kind of problem.

Try it yourself

Ready to try a couple on your own? Don't worry about getting the perfect answer right away. Focus on setting up the problem correctly using the principle of energy conservation.

Problem 1: A 2.0 kg mass is attached to a spring and oscillates with a total mechanical energy of 4.0 J. If the spring constant is 200 N/m, what is the amplitude of the motion?

Hint: At the amplitude, what form does all the energy take? Use the equation that relates total energy to amplitude.

Problem 2: An object is oscillating in SHM. At what position, in terms of its amplitude A, is its kinetic energy exactly equal to its potential energy?

Hint: Start with the conservation of energy equation, E_total = K + U. If K = U, what can you say about U in relation to E_total? Then substitute the formulas for E_total and U.

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