Unit 7.4 | Energy in Simple Harmonic Motion
- 14 minutes
- 2 formulas
- 16 questions
- 3 videos
Kinetic at the middle, potential at the ends, the same total throughout. This lesson uses energy to find speeds and amplitudes in the four SHM energy problem types.
Unit 7 Breakdown
In this lesson:
- Track the energy of an oscillator: kinetic at the middle, potential at the ends, the same total throughout
- Total energy from the amplitude, [katex] E = \tfrac{1}{2}kA^2 [/katex], and the speed at any position
- Read energy vs. position and energy vs. time graphs
- The 4 problem types: total energy and top speed, speed at a position, energy graphs, and changing the system mid-cycle
Introduction
This lesson covers energy in simple harmonic motion: how an oscillator trades between kinetic and potential energy every cycle, and how to use that trade to find speeds and amplitudes. It is the last lesson of Unit 7.
Nothing here is new physics. It is conservation of energy from Unit 3, applied to something that moves back and forth instead of sliding down a hill. That makes it a shortcut: the energy route gives speeds that the force route cannot reach without calculus.
Energy Around the Cycle
To see energy in simple harmonic motion, take a block on a horizontal spring with no friction. The spring and block form a closed system, so its total mechanical energy is constant:
[katex] E = K + U_s = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex]
- At a turning point ([katex] x = \pm A [/katex]), the block is at rest: all the energy is in the spring, [katex] E = \tfrac{1}{2}kA^2 [/katex].
- At equilibrium ([katex] x = 0 [/katex]), the spring is relaxed: all the energy is kinetic, [katex] E = \tfrac{1}{2}mv_{max}^2 [/katex].
- In between, it is a mix, and the mix depends only on where the block is.
[katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}mv_{max}^2 [/katex]
A pendulum does the same trade with gravitational energy: all [katex] mgh [/katex] at the top of each swing, all [katex] \tfrac{1}{2}mv^2 [/katex] at the bottom.
The total energy depends on the amplitude squared. The period does not depend on the amplitude at all. Pull a spring twice as far and the oscillator has 4 times the energy and twice the top speed, but the same period.
Every SHM energy question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Total energy and top speed | “total energy”, “maximum speed”, the amplitude changes | [katex] E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv_{max}^2 [/katex] |
| 2. Speed at a position | “speed when x = …”, “where is K = U”, a pendulum’s speed at the bottom | [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex] |
| 3. Energy graphs | graphs of [katex] K [/katex], [katex] U [/katex] or [katex] E [/katex] against position or time | Parabolas against [katex] x [/katex]; against time they repeat every [katex] \frac{T}{2} [/katex] |
| 4. Changing the system mid-cycle | clay lands on the block, a block falls off, a collision | Momentum for the collision, then energy for the new amplitude |
Problem Type 1: Total energy and top speed
How to spot it: The question asks for the total energy, the maximum speed, or what happens to them when the amplitude, mass or spring changes.
Setup: At a turning point all the energy is in the spring: [katex] E = \tfrac{1}{2}kA^2 [/katex]. At equilibrium it is all kinetic, so [katex] v_{max} = \sqrt{\frac{2E}{m}} = A\sqrt{\frac{k}{m}} [/katex].
A [katex] 0.50\,\text{kg} [/katex] block on a spring with [katex] k = 200 \text{ N/m} [/katex] oscillates with an amplitude of [katex] 0.10\,\text{m} [/katex] on a frictionless surface. Find its total energy and its maximum speed. What happens to both if the amplitude is doubled?
[katex] E = \tfrac{1}{2}(200)(0.10)^2 = 1.0 \text{ J} [/katex]. [katex] v_{max} = \sqrt{\frac{2(1.0)}{0.50}} = 2.0 \text{ m/s} [/katex]. Doubling the amplitude: [katex] E = \tfrac{1}{2}(200)(0.20)^2 = 4.0 \text{ J} [/katex] (4 times as much) and [katex] v_{max} = 4.0 \text{ m/s} [/katex] (twice as fast). The period stays [katex] 0.31\,\text{s} [/katex].
AP trap: Saying a bigger amplitude means a longer period because the block has farther to go. It also moves faster, by exactly enough to keep the period the same. Only the energy and the top speed change.
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Problem Type 2: Speed at a position
How to spot it: The question asks how fast the object moves at a given displacement, where the kinetic and potential energies are equal, or how fast a pendulum moves at the bottom.
Setup: Write the energy at the turning point and at the position you want: [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex]. For a pendulum, use [katex] mgh = \tfrac{1}{2}mv^2 [/katex] with [katex] h = L(1 – \cos\theta) [/katex].
The same block ([katex] 0.50\,\text{kg} [/katex], [katex] k = 200 \text{ N/m} [/katex], [katex] A = 0.10 \text{ m} [/katex]). How fast is it moving at [katex] x = 0.060 \text{ m} [/katex]? At what position are its kinetic and potential energies equal?
At [katex] 0.060\,\text{m} [/katex] the spring holds [katex] \tfrac{1}{2}(200)(0.060)^2 = 0.36 \text{ J} [/katex], so [katex] K = 1.0 – 0.36 = 0.64 \text{ J} [/katex] and [katex] v = \sqrt{\frac{2(0.64)}{0.50}} = 1.6 \text{ m/s} [/katex].
Equal energies means [katex] U_s = \tfrac{1}{2}E [/katex]: [katex] \tfrac{1}{2}kx^2 = \tfrac{1}{2}\left(\tfrac{1}{2}kA^2\right) [/katex], so [katex] x = \frac{A}{\sqrt{2}} \approx 0.071 \text{ m} [/katex]. Not halfway: at [katex] \frac{A}{2} [/katex] the spring holds only a quarter of the energy.
A pendulum [katex] 1.0\,\text{m} [/katex] long is released from rest at [katex] 10^\circ [/katex]. How fast is the bob moving at the bottom?
It drops [katex] h = L(1 – \cos\theta) = (1.0)(1 – \cos 10^\circ) = 0.015 \text{ m} [/katex]. Then [katex] v = \sqrt{2gh} = \sqrt{2(9.8)(0.0152)} \approx 0.55 \text{ m/s} [/katex]. The mass cancels, just like it does in the period.
AP trap: Assuming the kinetic and potential energies are equal halfway to the turning point. [katex] U_s [/katex] grows with [katex] x^2 [/katex], so at [katex] x = \frac{A}{2} [/katex] the spring has [katex] \frac{1}{4} [/katex] of the energy and the block has [katex] \frac{3}{4} [/katex].
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Problem Type 3: Energy graphs
How to spot it: A graph of kinetic, potential or total energy against position or against time, or a question that asks you to pick or sketch one.
Setup: Against position: [katex] U_s [/katex] is an upward parabola with its bottom at [katex] x = 0 [/katex], [katex] K [/katex] is the upside-down parabola, and [katex] E [/katex] is a flat line. Against time: [katex] U_s [/katex] and [katex] K [/katex] each peak twice per cycle, so they repeat every [katex] \frac{T}{2} [/katex].
Energy against position
Energy against time
A graph of a block’s spring potential energy against position is a parabola that reaches [katex] 0.50\,\text{J} [/katex] at [katex] x = \pm 0.10 \text{ m} [/katex], where the block turns around. The block has mass [katex] 0.25\,\text{kg} [/katex]. Find the spring constant and the block’s speed at [katex] x = 0.050 \text{ m} [/katex].
The turning points are at [katex] \pm 0.10 \text{ m} [/katex], so [katex] A = 0.10 \text{ m} [/katex] and [katex] E = 0.50 \text{ J} [/katex]. [katex] k = \frac{2E}{A^2} = \frac{2(0.50)}{(0.10)^2} = 100 \text{ N/m} [/katex]. At [katex] 0.050\,\text{m} [/katex]: [katex] U_s = \tfrac{1}{2}(100)(0.050)^2 = 0.125 \text{ J} [/katex], [katex] K = 0.375 \text{ J} [/katex], [katex] v = \sqrt{\frac{2(0.375)}{0.25}} \approx 1.7 \text{ m/s} [/katex].
AP trap: Giving an energy vs. time graph the same period as the motion. Kinetic energy peaks every time the block passes through the middle, which happens twice per cycle, so the energy graphs repeat every half period.
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Problem Type 4: Changing the system mid-cycle
How to spot it: Something lands on, sticks to, falls off or collides with the oscillating object, and the question asks about the new amplitude, top speed, energy or period.
Setup: Split it at the event. During a sticky collision, momentum is conserved but kinetic energy is not. After it, energy is conserved again: find the new total energy, then [katex] A = \sqrt{\frac{2E}{k}} [/katex]. The new period comes from the new mass.
A [katex] 0.50\,\text{kg} [/katex] block oscillates on a spring with [katex] k = 200 \text{ N/m} [/katex] and amplitude [katex] 0.10\,\text{m} [/katex], so it passes through equilibrium at [katex] 2.0\,\text{m/s} [/katex]. (a) As it passes through equilibrium, a [katex] 0.50\,\text{kg} [/katex] lump of clay dropped straight down lands on it and sticks. (b) Instead, the clay lands on it at a turning point. In each case, find the new amplitude and period.
(a) Horizontal momentum is conserved in the collision: [katex] (0.50)(2.0) = (1.0)v [/katex], so [katex] v = 1.0 \text{ m/s} [/katex]. The new energy is [katex] \tfrac{1}{2}(1.0)(1.0)^2 = 0.50 \text{ J} [/katex], half of the [katex] 1.0\,\text{J} [/katex] it had, so [katex] A = \sqrt{\frac{2(0.50)}{200}} \approx 0.071 \text{ m} [/katex].
(b) At a turning point the block is at rest, so the clay lands on a block with no kinetic energy to lose. The energy stays [katex] 1.0\,\text{J} [/katex], all in the spring, and the amplitude stays [katex] 0.10\,\text{m} [/katex]. The top speed drops to [katex] \sqrt{\frac{2(1.0)}{1.0}} \approx 1.4 \text{ m/s} [/katex].
In both cases the mass doubles, so the period grows from [katex] 2\pi\sqrt{\frac{0.50}{200}} \approx 0.31 \text{ s} [/katex] to [katex] 2\pi\sqrt{\frac{1.0}{200}} \approx 0.44 \text{ s} [/katex].
AP trap: Using conservation of energy through a sticky collision. Kinetic energy is lost when the clay sticks, so use momentum for the collision and energy only before and after it.
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Unit 7 Wrap-Up
- SHM comes from a restoring force proportional to the displacement: [katex] F = -kx [/katex]. Force and acceleration are largest at the turning points, speed at equilibrium (7.1).
- Period: [katex] T_s = 2\pi\sqrt{\frac{m}{k}} [/katex] and [katex] T_p = 2\pi\sqrt{\frac{L}{g}} [/katex], neither one depends on the amplitude (7.2).
- Graphs and equations: [katex] x = A\cos(2\pi f t) [/katex]; velocity is the slope of position, acceleration is position flipped (7.3).
- Energy: [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex] at every point of the cycle (7.4).
For the whole unit on one page, see the Simple Harmonic Motion Speed Review, or watch the SHM section of my speed review:
Ready for the whole unit? Take the unit review:
Unit 8 Preview
Next up is Unit 8: Fluids. Density, pressure, buoyancy and flow: the same forces and energy you know, applied to water and air.
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This article was written by Jason