Springs, rubber bands, and deformable solids store energy when stretched or compressed. This energy is elastic potential energy, and it obeys a remarkably simple law: \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the deformation. This section develops elastic potential energy from Hooke's law, explores the harmonic oscillator (the simplest yet most fundamental oscillating system in physics), and shows how elastic and gravitational energy combine in real systems.
Hooke's Law and spring force
A spring exerts a restoring force proportional to how far it is displaced from equilibrium:
\[ F_{\text{spring}} = -kx \]
where x is the displacement from the equilibrium position and k is the spring constant [N/m]. The negative sign indicates that the force opposes the displacement — if you pull the spring right (\( x > 0 \)), it pulls back left (\( F < 0 \)).
Physical meaning of \( k \):
- Large \( k \) (stiff spring): large force for small displacement
- Small \( k \) (soft spring): small force for large displacement
Work done to compress a spring
When you compress a spring from \( x = 0 \) to \( x = x_0 \), you do work against the spring force. The force you apply is \( F_{\text{applied}} = kx \) (positive, in the direction of motion). The work is:
\[ W_{\text{applied}} = \int_0^{x_0} kx \, dx = \frac{1}{2}k x_0^2 \]
This work is stored as potential energy in the spring. Therefore:
\[ U_{\text{spring}}(x) = \frac{1}{2}kx^2 \]
(with the zero-point at the equilibrium position, \( U(0) = 0 \)).
Force recovery from elastic PE
From §8.1, we recover the force via:
\[ F = -\frac{dU}{dx} = -\frac{d}{dx}\left(\frac{1}{2}kx^2\right) = -kx \]
Correct: Hooke's law. The elastic potential energy encodes the spring's behavior completely.
The harmonic oscillator
The harmonic oscillator is any system governed by \( F = -kx \). It's the simplest model of oscillation and appears everywhere: springs, pendulums (for small angles), vibrating atoms, electromagnetic waves, and quantum systems.
Equation of motion
Applying Newton's second law to a mass \( m \) attached to a spring:
\[ F = ma = -kx \]
\[ m\frac{d^2x}{dt^2} = -kx \]
\[ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \]
Define the angular frequency:
\[ \omega = \sqrt{\frac{k}{m}} \]
Then the equation becomes:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
Solution: The general solution is:
\[ x(t) = A\cos(\omega t + \phi) \]
where A is the amplitude and \phi is the phase. The system oscillates at a fixed frequency \( f = \omega / (2\pi) \), independent of amplitude (as long as Hooke's law holds).
Energy in a harmonic oscillator
At any instant, the total mechanical energy is:
\[ E = KE + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant} \]
At maximum displacement (\( x = A \), \( v = 0 \)):
\[ E = 0 + \frac{1}{2}kA^2 = \frac{1}{2}kA^2 \]
At equilibrium (\( x = 0 \), \( v = v_{\max} \)):
\[ E = \frac{1}{2}mv_{\max}^2 + 0 = \frac{1}{2}kA^2 \]
From the second equation: \( v_{\max} = A\omega = A\sqrt{k/m} \).
Energy oscillates between kinetic and potential, but the total is conserved.
Example: Mass on a spring
A \( 0.5 \) kg mass is attached to a spring with \( k = 200 \) N/m. The mass is pulled \( 0.1 \) m from equilibrium and released. What is the amplitude of oscillation, the maximum velocity, and the total energy?
Amplitude: \( A = 0.1 \) m (the maximum displacement is the amplitude).
Angular frequency: \( \omega = \sqrt{k/m} = \sqrt{200/0.5} = \sqrt{400} = 20 \) rad/s
Maximum velocity: \( v_{\max} = A\omega = 0.1 \times 20 = 2 \) m/s
Total energy: \( E = \frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \) J
This 1 J of energy oscillates continuously between kinetic form (when the mass passes through equilibrium at 2 m/s) and potential form (when the mass momentarily stops at the extreme positions).
Combining gravity and elastic PE: vertical spring
When a spring hangs vertically and a mass is attached, both gravity and elastic PE contribute. Let \( y \) be the vertical position (positive upward) and \( y_0 \) be the equilibrium position (where \( kx_0 = mg \), with \( x_0 \) the stretch).
Total potential energy:
\[ U_{\text{total}}(y) = mgy + \frac{1}{2}k(y_0 - y)^2 \]
The first term is gravitational PE; the second is elastic PE (measuring stretch from the natural length at \( y_0 \)).
Simplification: By shifting coordinates to measure displacement from equilibrium, we can show that oscillations about equilibrium are still governed by \( \frac{1}{2}k\xi^2 \), where \( \xi \) is displacement from the new equilibrium. Gravity merely shifts the equilibrium position but doesn't change the oscillation frequency.
Elastic PE in real systems
Example: Bungee jumping
A jumper of mass \( m = 75 \) kg leaps from a bridge. The bungee cord has spring constant \( k = 50 \) N/m and natural length \( L_0 = 20 \) m. The bridge is \( h = 50 \) m above a safety platform. What is the maximum stretch of the cord?
Energy conservation: At the lowest point, all gravitational PE is converted to elastic PE (assuming air resistance is negligible).
From the initial height to maximum stretch \( x_{\max} \):
\[ \text{Initial PE (gravity)} = \text{Final PE (elastic)} \]
More precisely:
\[ mg(L_0 + x_{\max}) = \frac{1}{2}kx_{\max}^2 \]
(The jumper falls a total distance \( L_0 + x_{\max} \) to reach the lowest point; gravity does work \( mg(L_0 + x_{\max}) \).)
Rearranging:
\[ \frac{1}{2}kx_{\max}^2 - mg x_{\max} - mg L_0 = 0 \]
\[ \frac{1}{2} \times 50 \times x_{\max}^2 - 75 \times 9.8 \times x_{\max} - 75 \times 9.8 \times 20 = 0 \]
\[ 25 x_{\max}^2 - 735 x_{\max} - 14{,}700 = 0 \]
Using the quadratic formula:
\[ x_{\max} = \frac{735 \pm \sqrt{735^2 + 4 \times 25 \times 14{,}700}}{2 \times 25} = \frac{735 \pm \sqrt{540{,}225 + 1{,}470{,}000}}{50} \]
\[ = \frac{735 \pm \sqrt{2{,}010{,}225}} \approx \frac{735 \pm 1{,}418}{50} \]
Taking the positive root: \( x_{\max} \approx \frac{2{,}153}{50} \approx 43 \) m
The cord stretches about 43 m, so the jumper descends a total of \( 20 + 43 = 63 \) m, which exceeds the bridge height of 50 m — they hit the platform! This example shows the danger of miscalculating spring constants or jump heights.
Connection to Chapter 7 (Work and Energy)
§7.2 introduced work by a variable force and showed that \( W = \int F \, dx = \frac{1}{2}kx^2 \) for a spring. §8.3 now places this into the potential energy framework: the work done by the spring equals the negative change in elastic PE, \( W = -\Delta U \), which is exactly what we found by integration.
What you should be able to do
After this section, you should be able to:
- Apply Hooke's law \( F = -kx \) to find the force exerted by a spring
- Derive and use elastic potential energy \( U = \frac{1}{2}kx^2 \)
- Solve the equation of motion for a harmonic oscillator and identify amplitude, frequency, and phase
- Use energy conservation to find the maximum velocity or maximum displacement of an oscillating mass
- Combine gravitational and elastic PE to analyze systems like vertical springs or bungee jumps
- Explain why gravity shifts the equilibrium position but doesn't change the frequency of oscillation
Preview of §8.4: Mechanical Energy Conservation
When only conservative forces (gravity, springs, etc.) act on a system, the total mechanical energy is conserved. §8.4 develops this principle systematically, derives the condition for energy conservation, and explores what happens when we add non-conservative forces like friction.
📚 See also: Halliday Vol 1, Ch 8, §8.3 — Elastic Potential Energy and Simple Harmonic Motion. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 8.4: Elastic Potential Energy and Simple Harmonic Oscillation. 🎓 Video: MIT 8.01 Lecture 13 (Walter Lewin) — "Simple Harmonic Oscillator".
Have a question? 🤔
If something isn't clear or you have a question, ask it here. The answer will be published on this page.