Difficulty: Intermediate | Prerequisites: basic algebra, unit conversions (rpm to Hz, grams to kg), familiarity with trigonometric functions.
This material sits at the heart of the first midterm for PHYS 116 at UW. Simple harmonic motion (SHM) is the gateway to understanding waves, sound, and eventually optics later in the course. If you can describe how a mass on a spring or a pendulum moves, stores energy, and loses energy to damping, every wave topic that follows will make more sense. You should already be comfortable with Newton's second law, conservation of energy, and basic kinematics.
SHM describes any system where the restoring force is proportional to displacement. The key quantities (period, frequency, amplitude, angular frequency) are all interrelated and show up in every problem. Energy in SHM swaps between kinetic and potential forms, but the total stays constant unless damping is present, in which case amplitude decays exponentially.
Period (T)
The time for one complete oscillation, measured in seconds. Think of it as how long the system takes to return to where it started.
Frequency (f)
The number of complete oscillations per second, measured in hertz (Hz). It is the reciprocal of the period: f = 1/T. In simple terms, frequency tells you how many cycles happen each second.
Angular frequency (ω)
The rate of oscillation expressed in radians per second: ω = 2πf = 2π/T. It appears in the position, velocity, and acceleration equations for SHM.
Amplitude (A)
The maximum displacement from the equilibrium position. It sets the scale of the motion but does not affect the period or frequency in ideal SHM.
Spring constant (k)
A measure of a spring's stiffness, in N/m. A stiffer spring (larger k) produces a shorter period for the same mass.
Linear mass density (μ)
Mass per unit length of a string or spring, in kg/m. It determines wave speed on a string alongside tension.
Damping 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 the system loses energy more slowly.
Equilibrium position
The position where the net force on the oscillating object is zero. For a horizontal spring, this is the natural (unstretched) length.
Converting rpm to Hz: divide by 60. An engine at 3200 rpm runs at 3200/60 = 53.3 Hz.
Converting period to frequency: f = 1/T. A fire alarm with T = 2.4 × 10⁻³ s has f = 417 Hz.
These conversions appear in nearly every problem. Get them wrong and the rest of the calculation collapses.
Position: x(t) = A cos(ωt + φ)
Velocity: v(t) = –Aω sin(ωt + φ)
Acceleration: a(t) = –Aω² cos(ωt + φ) = –ω² x(t)
Maximum positive velocity occurs when the object passes through equilibrium moving in the positive direction. On an acceleration-vs-time graph, this is where a(t) crosses zero going from positive to negative.
Maximum positive displacement corresponds to the most negative acceleration (a = –ω²A).
Total mechanical energy: E = (1/2)kA², constant throughout the motion.
At any position x: E = (1/2)kx² + (1/2)mv².
To find amplitude from a known position and velocity, rearrange: A = √(x² + mv²/k).
Maximum kinetic energy equals the total energy and occurs at x = 0: K_max = (1/2)mA²ω².
From a position equation like x = (0.04 m) cos(15.0t), read off A = 0.04 m and ω = 15.0 rad/s directly, then compute K_max.
For two systems with the same mass m and same amplitude A but different spring constants (k vs 4k):
Period: T = 2π√(m/k), so the stiffer spring (4k) gives half the period.
Maximum potential energy: U_max = (1/2)kA², so the 4k system stores four times the energy.
Maximum kinetic energy equals maximum potential energy, so K_max is also four times larger.
Maximum speed: since K_max = (1/2)mv²_max, the 4k system has twice the maximum speed.
If a second block of mass m is attached at the turning point (x = ±A, where velocity is zero), the amplitude does not change because no kinetic energy is added or removed.
The spring constant stays the same, so maximum potential energy is unchanged.
Therefore maximum kinetic energy is equal to the original system's.
The period does increase because T = 2π√(m_total/k).
Period: T = 2π√(l/g), valid for small angles (typically under 15°).
The period does not depend on mass or amplitude (for small swings).
To find length from observed oscillations: measure total time for N cycles, compute T = t_total/N, then solve l = T²g/(4π²).
Amplitude decays exponentially: A(t) = A₀ e^(–t/τ), where τ is the damping time constant.
To find the time to reach a fraction of the initial amplitude, set A/A₀ = fraction, take the natural log, and solve: t = –τ ln(A/A₀).
For 70% of initial amplitude: t = –τ ln(0.7) = –(45.0 s)(–0.357) = 16.1 s.
f = \frac{1}{T}\omega = 2\pi f = \frac{2\pi}{T}x(t) = A \cos(\omega t + \phi)v_{max} = A\omegaE_{total} = \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2A = \sqrt{x^2 + \frac{mv^2}{k}}K_{max} = \frac{1}{2}mA^2\omega^2T_{spring} = 2\pi\sqrt{\frac{m}{k}}T_{pendulum} = 2\pi\sqrt{\frac{l}{g}}A(t) = A_0 \, e^{-t/\tau}t = -\tau \ln\!\left(\frac{A}{A_0}\right)Car suspensions are damped spring-mass systems, designed so that the damping time constant gives a smooth ride without excessive bouncing. The same maths that predicts how a block on a spring loses amplitude describes how a car settles after hitting a bump.
Pendulum clocks rely on the period of a simple pendulum being independent of amplitude (for small swings). Clockmakers adjust the pendulum length to tune the period to exactly one second per half-swing.
Students often think maximum velocity occurs at maximum displacement. It does not. Maximum velocity occurs at the equilibrium position (x = 0), where all the energy is kinetic.
Students confuse angular frequency (ω, in rad/s) with frequency (f, in Hz). They differ by a factor of 2π, and using one where the other belongs will throw off your answer.
Students sometimes assume that adding mass to an oscillating system changes the amplitude. If the mass is added at a turning point (v = 0), the amplitude stays the same because no kinetic energy is added or removed.
Students forget to convert grams to kilograms before plugging into formulas. A 200 g ball is 0.200 kg, and using 200 in place of 0.200 gives answers off by a factor of 1000.
⚠️ Unit conversions (rpm to Hz, period to frequency, grams to kg) are tested in nearly every problem. Get comfortable doing these without thinking.
⚠️ Reading ω and A directly from a position equation like x = A cos(ωt) is a common exam move. Know which number is which.
⚠️ Energy conservation problems will give you a position and velocity at one point and ask for the amplitude. You must use E = (1/2)kx² + (1/2)mv² = (1/2)kA².
⚠️ Pendulum problems often give total time for N oscillations. Divide to get the period of one oscillation before doing anything else.
⚠️ Damping problems test whether you can manipulate the exponential decay equation. Know how to take the natural log and solve for time.
True or false: the period of a simple pendulum depends on the mass of the bob. (False)
Fill in the blank: maximum speed in SHM occurs at the ______ position. (equilibrium)
True or false: doubling the spring constant while keeping mass and amplitude the same doubles the maximum kinetic energy. (False, it quadruples it)
Fill in the blank: to convert rpm to Hz, divide by ______. (60)
True or false: if a second block is attached at x = –A (where v = 0), the amplitude changes. (False)
Q: A car engine runs at 3200 rpm and a fire alarm has a period of 2.4 × 10⁻³ s. A string oscillates at (f_alarm – f_engine)/2. What is the string's frequency?
A: f_engine = 3200/60 = 53.3 Hz. f_alarm = 1/(2.4 × 10⁻³) = 417 Hz. String frequency = (417 – 53.3)/2 = 182 Hz. Closest answer: 180 Hz.
Q: A 200 g ball on a spring (k = 3.50 N/m) has velocity 0.300 m/s at x = –5.00 cm. What is the amplitude?
A: A = √(x² + mv²/k) = √((0.05)² + (0.2)(0.3)²/3.5) = √(0.0025 + 0.00514) = 0.0874 m = 8.74 cm.
Q: The position of a 70.0 g mass on a spring is x = (0.0400 m) cos(15.0t). What is the maximum kinetic energy?
A: K_max = (1/2)m A² ω² = (1/2)(0.07)(0.04)²(15)² = 0.0126 J = 12.6 × 10⁻³ J.
Q: A 200 g ball on a string completes 20 oscillations in 17.0 s. How long is the string?
A: T = 17.0/20 = 0.85 s. l = T²g/(4π²) = (0.85)²(9.8)/(4π²) = 0.179 m.
Q: A damped oscillator has τ = 45.0 s. How long until the amplitude drops to 70% of its initial value?
A: t = –τ ln(0.7) = –(45.0)(–0.357) = 16.1 s.
Q: Block D (spring constant k) and Block E (spring constant 4k) have the same mass and amplitude. Is the period of D greater than, less than, or equal to the period of E?
A: Greater than. T = 2π√(m/k), so the smaller spring constant gives a longer period.
SHM is the building block for understanding transverse and longitudinal waves. The same sinusoidal functions that describe a mass on a spring also describe the displacement of a point on a vibrating string or a sound wave in air.
Energy conservation in SHM connects directly to the work-energy theorem from earlier in the course. The trade-off between kinetic and potential energy here reappears in wave energy and sound intensity problems.
Damped oscillations lead into driven oscillations and resonance, which explain why buildings sway in earthquakes and why opera singers can shatter glass.
Simple harmonic motion, SHM, oscillation, spring-mass system, Hooke's law, restoring force, period, frequency, angular frequency, amplitude, phase constant, energy conservation in SHM, kinetic energy, potential energy, spring constant, pendulum, simple pendulum, small-angle approximation, damped oscillation, damping time constant, exponential decay, PHYS 116, University of Washington, general physics, midterm review