Difficulty: Intermediate | Prerequisites: PHYS 115 or equivalent (Newton's laws, energy conservation, basic trigonometry)
Oscillations and simple harmonic motion (SHM) form the bridge between mechanics and wave physics. Nearly every vibrating system, from a child's swing to the atoms in a crystal lattice, can be modelled as a harmonic oscillator once the displacement is small enough. This unit introduces the mathematics of periodic motion and connects it to energy conservation, which you covered in PHYS 115. If you can describe a mass on a spring, you already have the toolkit for understanding sound waves, light, and AC circuits later in the course.
Any system that experiences a restoring force proportional to displacement will oscillate sinusoidally. The period of a mass-spring system depends on mass and spring constant, while a pendulum's period depends on length and gravity. Energy swaps continuously between kinetic and potential, and total mechanical energy is set by the amplitude.
Simple harmonic motion (SHM)
Motion in which the restoring force is directly proportional to the displacement from equilibrium and directed toward the equilibrium point. In simple terms, the further you pull it, the harder it pulls back, and it overshoots every time.
Amplitude (A)
The maximum displacement from the equilibrium position, measured in metres. Think of it as how far the oscillator swings from the centre at its most extreme.
Period (T)
The time for one complete oscillation cycle, measured in seconds. One full back-and-forth counts as one period.
Frequency (f)
The number of complete oscillation cycles per second, measured in hertz (Hz). It is the reciprocal of the period: f = 1/T.
Angular frequency (omega)
The rate of oscillation measured in radians per second: omega = 2 pi f. It appears in every SHM equation and connects the oscillation to circular motion.
Spring constant (k)
A measure of a spring's stiffness, in N/m. A larger k means a stiffer spring that exerts more force per unit of stretch.
Equilibrium position
The position where the net force on the oscillator is zero. The object naturally sits here when nothing disturbs it.
Damping time constant (tau)
The time for the amplitude of a damped oscillator to fall to 1/e (about 37%) of its initial value. A larger tau means the oscillation dies out more slowly.
Phase constant (phi)
The angle (in radians) that sets where in its cycle the oscillator starts at t = 0. It tells you whether the object begins at the amplitude, at equilibrium, or somewhere in between.
Position as a function of time: x(t) = A cos(omega t + phi)
Velocity: v(t) = -A omega sin(omega t + phi). Maximum speed v_max = A omega, occurring at the equilibrium position.
Acceleration: a(t) = -A omega^2 cos(omega t + phi) = -omega^2 x. Maximum acceleration a_max = A omega^2, occurring at the turning points.
The acceleration is always opposite in sign to the displacement. When the graph of acceleration versus time crosses zero going negative, the velocity is at its maximum positive value (the object is passing through equilibrium heading in the positive direction). When the acceleration is at its most negative, the object is at its maximum positive position.
f = 1/T and T = 1/f
To convert revolutions per minute to Hz: divide by 60.
Example: 3200 rpm = 3200/60 = 53.3 Hz
To find frequency from period: f = 1/T.
Example: T = 2.4 x 10^-3 s gives f = 1/(2.4 x 10^-3) = 416.7 Hz
Total mechanical energy is conserved in an ideal (undamped) oscillator:
E_total = (1/2) k A^2
At any position x: E_total = (1/2) k x^2 + (1/2) m v^2
At the turning points (x = plus or minus A): all energy is potential, KE = 0.
At equilibrium (x = 0): all energy is kinetic, KE_max = (1/2) k A^2.
To find the amplitude when you know the velocity at a particular position, rearrange the energy equation:
(1/2) k A^2 = (1/2) k x^2 + (1/2) m v^2
A = sqrt(x^2 + (m/k) v^2)
If x(t) = A cos(omega t), you can read off A and omega directly.
omega = 15.0 rad/s and A = 0.0400 m gives v_max = A omega = 0.600 m/s
KE_max = (1/2) m v_max^2
Example: m = 0.070 kg, v_max = 0.600 m/s gives KE_max = (1/2)(0.070)(0.600)^2 = 0.0126 J = 12.6 x 10^-3 J
For small angles (typically under about 15 degrees), a pendulum behaves as an SHM system.
Period: T = 2 pi sqrt(L / g)
The period does not depend on mass or amplitude (for small angles).
To find the length from timing data: measure the period by dividing total time by the number of oscillations, then solve for L.
Example: 20 oscillations in 17.0 s gives T = 0.850 s. Then L = g (T / 2 pi)^2 = 9.80 (0.850 / 6.283)^2 = 0.179 m.
In a damped oscillator, friction or drag removes energy from the system over time. The amplitude decays exponentially:
A(t) = A_0 e^(-t / 2 tau)
Here tau is the damping time constant. Note that in some textbooks the exponent uses t/tau without the factor of 2, depending on whether tau is defined for amplitude or for energy. In PHYS 116, the convention is:
A(t) = A_0 e^(-t / 2 tau)
To find the time for the amplitude to fall to a fraction f of A_0: set f = e^(-t / 2 tau) and solve.
t = -2 tau ln(f)
Example: tau = 45.0 s, f = 0.70. t = -2 (45.0) ln(0.70) = -90.0 (-0.357) = 16.1 s.
Consider two experiments: Block D on a spring with constant k, and Block E on a spring with constant 4k. Both have the same mass m and oscillate between +A and -A.
Period of D: T_D = 2 pi sqrt(m/k)
Period of E: T_E = 2 pi sqrt(m/4k) = T_D / 2
Block D has a greater period (slower oscillation) than Block E, because the stiffer spring produces a larger restoring force and drives the mass through its cycle faster.
KE_max for D: (1/2) k A^2
KE_max for E: (1/2)(4k) A^2 = 4 times (1/2) k A^2
The stiffer spring stores four times as much total energy for the same amplitude. The KE-versus-position curve for E is a parabola that peaks four times higher than D's curve.
v_max for D: A omega_D = A sqrt(k/m)
v_max for E: A omega_E = A sqrt(4k/m) = 2 A sqrt(k/m) = 2 v_max,D
Block E reaches twice the maximum speed and completes its cycle in half the time, so its velocity-versus-time graph is a sinusoid with twice the amplitude and twice the frequency of D's.
If a second block of mass m is attached to Block E at the instant it reaches -A (where v = 0), the amplitude does not change (the block is momentarily at rest, so no momentum is exchanged). The total energy remains (1/2)(4k) A^2.
The new combined block F has mass 2m on the same spring, so its maximum kinetic energy equals the same total energy: KE_max,F = (1/2)(4k) A^2 = KE_max,E. Adding mass at the turning point does not change the maximum KE.
x(t) = A \cos(\omega t + \phi)v(t) = -A\omega \sin(\omega t + \phi)a(t) = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x\omega = 2\pi f = \frac{2\pi}{T}T_{\text{spring}} = 2\pi \sqrt{\frac{m}{k}}T_{\text{pendulum}} = 2\pi \sqrt{\frac{L}{g}}E_{\text{total}} = \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2A(t) = A_0 \, e^{-t / 2\tau}Car engine vibrations are periodic oscillations measured in rpm. Engineers must account for resonance frequencies when designing engine mounts to avoid dangerous vibration buildup. The damping concepts from this unit are the reason shock absorbers in vehicles use specific damping constants: too little damping and the car bounces endlessly, too much and it transmits every bump directly.
Students often think a stiffer spring makes the period longer. It does not. A larger k means a shorter period because the restoring force is stronger and drives the mass through its cycle faster.
Students often confuse where maximum velocity occurs with where maximum acceleration occurs. Maximum velocity happens at the equilibrium position (x = 0). Maximum acceleration happens at the turning points (x = plus or minus A).
Students sometimes assume that adding mass to an oscillator at a turning point changes the amplitude. It does not, because the velocity is zero at the turning point and no kinetic energy is gained or lost.
Students often forget that the pendulum period formula only works for small angles. For large angles, the period becomes longer than 2 pi sqrt(L/g) predicts.
⚠️ Reading omega and A directly from a position equation like x = (0.0400 m) cos(15.0t) is tested repeatedly. Know how to extract both values at a glance.
⚠️ Energy conservation problems that give you velocity at a known position and ask for amplitude are a favourite. Set up (1/2)kA^2 = (1/2)kx^2 + (1/2)mv^2 and solve.
⚠️ Expect a question comparing two oscillators with different spring constants and the same amplitude. You need to track how period, v_max, and KE_max all change.
⚠️ Damped oscillation problems usually ask for the time to reach a given fraction of initial amplitude. Get comfortable with the natural log.
True or false: the period of a mass-spring oscillator depends on the amplitude of oscillation. (False.)
Fill in the blank: maximum kinetic energy of a spring-mass system equals ____. ((1/2) k A^2.)
True or false: doubling the spring constant while keeping mass and amplitude the same doubles the maximum speed. (True.)
Fill in the blank: for a simple pendulum, the period depends on ____ and ____ but not on ____. (Length and g, but not mass.)
True or false: in a damped oscillator, the amplitude decays linearly over time. (False, it decays exponentially.)
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: Use energy conservation. A = sqrt(x^2 + (m/k)v^2) = sqrt((0.0500)^2 + (0.200/3.50)(0.300)^2) = sqrt(0.0025 + 0.00514) = sqrt(0.00764) = 0.0874 m = 8.74 cm.
Q: The position of a 70.0 g mass is x = (0.0400 m) cos(15.0t). What is the maximum kinetic energy?
A: v_max = A omega = (0.0400)(15.0) = 0.600 m/s. KE_max = (1/2)(0.0700)(0.600)^2 = 12.6 x 10^-3 J.
Q: 20 oscillations of a pendulum take 17.0 s. How long is the string?
A: T = 17.0/20 = 0.850 s. L = g(T/2pi)^2 = (9.80)(0.850/6.283)^2 = 0.179 m.
Q: A damped oscillator has tau = 45.0 s. How long until the amplitude drops to 70% of its initial value?
A: t = -2 tau ln(0.70) = -2(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 D's period greater than, less than, or equal to E's?
A: Greater than. T = 2pi sqrt(m/k), so the smaller spring constant gives the longer period. T_D = 2 T_E.
SHM is the foundation for understanding travelling waves and standing waves, which are covered in the next part of PHYS 116. A travelling wave is a disturbance where each particle undergoes SHM as the wave passes. The energy concepts here (KE swapping with PE) carry directly into wave energy and sound intensity. Damping connects to real-world engineering problems and will resurface when you study driven oscillations and resonance.
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