Difficulty: Intermediate | Prerequisites: Sections 14.1 to 14.3 (SHM, springs, pendulums), energy conservation (Chapter 10)
Sections 14.1 to 14.3 established what SHM looks like. These sections explain what governs its energy, what happens when friction is present, and what happens when you push a system at its natural frequency. Energy methods give you a second, often faster, route to the same answers you can get from Newton's second law. Damping explains why real oscillations die out. Resonance explains why a wine glass shatters at a particular pitch, and why engineers worry about bridge vibrations. Together, 14.4 to 14.7 complete the oscillation toolkit you will need for waves in the next chapter.
In a frictionless mass-spring system, energy swaps continuously between kinetic and elastic potential, with the total staying constant. The frequency and period of a spring-mass oscillator depend only on k and m, while a pendulum's depend only on L (or I, m, d) and g. Real oscillations lose energy to friction (damping), and the amplitude decays exponentially. If you drive a damped oscillator at its natural frequency, the amplitude peaks sharply: that is resonance.
Elastic potential energy (U)
The energy stored in a spring when it is stretched or compressed from its natural length. U = ½kx². Think of it as the energy the spring "holds" when you pull or push it.
Total mechanical energy (E)
The sum of kinetic energy and potential energy: E = K + U = ½mv² + ½kx². In a frictionless system, E is constant throughout the motion. In simple terms, the total never changes; it just shifts between motion and spring storage.
Turning point
The position x = ±A where the object momentarily stops before reversing direction. All energy is potential at the turning points. Think of it as the ends of the swing, where everything is stored energy and nothing is motion.
Damped oscillation
An oscillation whose amplitude decreases over time because of a non-conservative force such as friction or air resistance (a drag force). In simple terms, a swing that gradually slows down and stops.
Time constant (τ)
The time it takes for the amplitude of a damped oscillator to fall to 1/e (about 37%) of its initial value. A larger τ means slower decay. Think of it as the "half-life-ish" number for how quickly oscillations die out.
Exponential decay
The amplitude envelope of a damped oscillator follows x_max(t) = Ae^(-t/τ), where A is the initial amplitude and τ is the time constant. In simple terms, the peaks get smaller by the same fraction in every equal time interval.
Driven oscillation
An oscillation sustained by an external periodic force. The system oscillates at the driving frequency, not necessarily its natural frequency. Think of it as pushing a child on a swing: you supply energy from outside.
Resonance
The condition where the driving frequency matches the system's natural frequency, producing a dramatically larger amplitude response. In simple terms, pushing at exactly the right rhythm to make the swing go as high as possible.
Natural frequency (f₀)
The frequency at which a system oscillates when displaced and released with no driving force or damping. For a spring: f₀ = (1/2π)√(k/m). Think of it as the system's preferred rhythm.
Energy conservation in a frictionless spring
Total mechanical energy E = K + U = ½mv² + ½kx² is constant throughout the motion.
At the turning points (x = ±A), velocity is zero, so K = 0 and all energy is potential: E = U_max = ½kA².
At the equilibrium position (x = 0), displacement is zero, so U = 0 and all energy is kinetic: E = K_max = ½mv_max².
Setting these equal: ½kA² = ½mv_max², which gives v_max = A√(k/m) = ωA.
Energy vs position
At any position x, kinetic energy is K = ½k(A² - x²) and potential energy is U = ½kx².
The energy-vs-position graph is a parabola for U and an inverted parabola for K, and the two always sum to the constant E.
The key insight
Higher spring constant k or larger amplitude A means more total energy.
But amplitude does not affect frequency or period. A system with a large amplitude oscillates with the same period as one with a small amplitude (same k and m). The object simply moves faster through the middle to cover the extra distance in the same time.
Mass-spring system
ω = √(k/m), so f = (1/2π)√(k/m) and T = 2π√(m/k).
A stiffer spring (higher k) gives a higher frequency and shorter period.
A heavier mass (higher m) gives a lower frequency and longer period.
Amplitude does not appear in any of these expressions. Period and frequency are set entirely by k and m.
Pendulum
f = (1/2π)√(g/L) and T = 2π√(L/g) for a simple pendulum.
A longer pendulum has a longer period. A shorter pendulum has a shorter period.
Pendulum frequency does not depend on mass.
On a planet with stronger gravity (larger g), a pendulum swings faster.
Physical pendulum
T = 2π√(I / (mgd)), where I is the moment of inertia about the pivot and d is the distance from the pivot to the centre of gravity.
For a uniform rod of length L pivoted at one end: I = (1/3)mL² and d = L/2.
Worked example from the notes
A pendulum has T = 7.9 s. Find L. T = 2π√(L/g), so L = gT²/(4π²) = 9.8 × 7.9² / (4π²) ≈ 15.5 m.
What is damping?
A damped oscillation is one that runs down and eventually stops because of a non-conservative force such as a drag force (air resistance, friction).
The system still oscillates, but the amplitude shrinks with each cycle.
Exponential decay of amplitude
The maximum displacement follows x_max(t) = Ae^(-t/τ), where A is the initial amplitude and τ (tau) is the time constant.
e is Euler's number, approximately 2.718.
After one time constant (t = τ), the amplitude has dropped to A/e ≈ 0.37A, roughly 37% of its starting value.
After 2τ, it is about 0.37² ≈ 14% of A. After 3τ, about 5%.
Reading a damping graph
The position-vs-time graph shows a sinusoidal oscillation whose peaks and troughs shrink inside a decaying exponential envelope.
Early on, the difference between successive peaks is large. Later, the peaks are closer together in height because the exponential curve flattens.
Worked example from the notes (pre-lecture Q1.C)
Initial amplitude A = 13.5 (given as x_max at t = 0). At a later time, x_max = 0.135. Then x_max/A = 0.135/13.5 = 0.01 = e^(-t/τ). Taking ln: -t/τ = ln(0.01) ≈ -4.6, so t ≈ 4.6τ.
Worked example (pre-lecture Q1.D)
If x_max(t) = ¾A, then e^(-t/τ) = 0.75. Taking ln: -t/τ = ln(0.75) ≈ -0.29, so t ≈ 0.29τ. With τ = 490 (from context), t ≈ 142 time units.
Driven oscillation basics
A driven (or forced) oscillator receives energy from an external periodic force at some driving frequency f_drive.
The system eventually settles into steady-state oscillation at f_drive, regardless of its natural frequency f₀.
The amplitude of the steady-state response depends on how close f_drive is to f₀.
Resonance
When f_drive = f₀ (the natural frequency), the amplitude of the response reaches a maximum. This condition is called resonance.
At resonance, the external force delivers energy in phase with the velocity, transferring energy as efficiently as possible.
With light damping, the resonance peak is tall and narrow: even a small mismatch between f_drive and f₀ causes the amplitude to drop sharply.
With heavy damping, the peak is shorter and broader: the system is less selective about frequency.
Why resonance matters
Resonance can be useful: musical instruments amplify sound at their resonant frequencies, and MRI scanners exploit nuclear magnetic resonance.
Resonance can be destructive: soldiers break step crossing a bridge to avoid driving it at its natural frequency, and the Tacoma Narrows Bridge collapse is a famous example of wind-driven resonance.
Energy in SHM
U = \tfrac{1}{2}kx^2E = K + U = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2 = \text{constant}v_{\max} = \omega A = A\sqrt{\frac{k}{m}}Frequency and period (mass-spring)
\omega = \sqrt{\frac{k}{m}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}Frequency and period (simple pendulum)
f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}, \qquad T = 2\pi\sqrt{\frac{L}{g}}Physical pendulum
T = 2\pi\sqrt{\frac{I}{mgd}}Damped oscillation
x_{\max}(t) = A\,e^{-t/\tau}After one time constant τ: x_max(τ) = A/e ≈ 0.37A.
Students often think total energy changes during SHM. In a frictionless system it does not. Energy shifts between kinetic and potential, but the sum is constant.
Students often believe amplitude affects period or frequency. It does not, for either springs or (small-angle) pendulums.
Students sometimes think damping changes the frequency of oscillation. For light damping the frequency stays very close to the natural frequency; only the amplitude decreases.
Students often confuse the driving frequency with the natural frequency. At resonance they are equal, but in general a driven oscillator oscillates at whatever frequency you drive it, not at its natural frequency.
⚠️ Energy problems are a common exam format: "Given k, m, and A, find v_max" or "At position x, what fraction of the energy is kinetic?" Know E = ½kA² and how to partition it.
⚠️ Expect a question comparing two systems with different k or m and asking which has the higher frequency. Remember: higher k or lower m means higher f.
⚠️ Damping questions typically give you A and ask for the amplitude after a certain number of time constants, or give you a ratio and ask you to solve for t/τ using logarithms.
⚠️ Resonance is often tested conceptually: what happens to amplitude when driving frequency equals natural frequency? What happens if damping increases?
True or false: At the turning points of SHM, all energy is kinetic. (False, it is all potential.)
Fill in the blank: The total energy of a mass-spring system is E = ½k______². (A, the amplitude)
True or false: A damped oscillator's amplitude decays linearly with time. (False, it decays exponentially.)
Fill in the blank: Resonance occurs when the driving frequency equals the ______ frequency. (natural)
True or false: Increasing damping makes the resonance peak taller and narrower. (False, it makes it shorter and broader.)
Q: A 0.5 kg block on a spring (k = 200 N/m) is pulled to x = 0.1 m and released. What is the total energy?
A: E = ½kA² = ½ × 200 × 0.1² = 1.0 J.
Q: In the same system, what is the maximum speed of the block?
A: v_max = A√(k/m) = 0.1 × √(200/0.5) = 0.1 × 20 = 2.0 m/s.
Q: At x = 0.05 m, what is the kinetic energy?
A: U = ½kx² = ½ × 200 × 0.05² = 0.25 J. K = E - U = 1.0 - 0.25 = 0.75 J.
Q: A damped oscillator starts with amplitude 10 cm and has a time constant τ = 4 s. What is the amplitude at t = 4 s?
A: x_max(4) = 10 × e^(-4/4) = 10 × e^(-1) ≈ 10 × 0.37 = 3.7 cm.
Q: A child pushes a swing at its natural frequency. What happens to the amplitude over time compared to pushing at a different frequency?
A: Pushing at the natural frequency (resonance) produces the largest possible amplitude. Pushing at any other frequency gives a smaller steady-state amplitude.
Energy conservation in SHM is a specific case of the broader work-energy theorem from Chapter 10. Damped oscillations connect to the concept of dissipative forces and thermal energy. Resonance reappears in wave physics (Chapter 15, standing waves), in electrical circuits (LC and RLC circuits), and in modern physics (spectroscopy, NMR). Understanding exponential decay also prepares you for radioactive decay in nuclear physics.
Elastic potential energy, kinetic energy, energy conservation, total mechanical energy, turning point, maximum speed, v_max, spring constant, mass-spring frequency, pendulum period, physical pendulum, moment of inertia, damped oscillation, damping, drag force, exponential decay, time constant, tau, amplitude decay, driven oscillation, forced oscillation, resonance, natural frequency, driving frequency, resonance peak, Tacoma Narrows Bridge, PHYS 101, Chapter 14, University of Washington General Physics