Source: Final Exam Formula Booklet
Tags: simple harmonic motion, SHM, periodic motion, pendulum, spring, torsion, amplitude, angular frequency, phase constant, ENGR 216, PHYS 216, Texas A&M
Simple harmonic motion (SHM) describes any system where the restoring force (or torque) is proportional to displacement. The position, velocity, and acceleration all follow sinusoidal functions. The three classic examples on this exam are the pendulum, the mass-spring system, and the torsional oscillator, each with its own period formula.
Simple harmonic motion (SHM)
Oscillatory motion where the acceleration is always proportional to, and directed opposite to, the displacement from equilibrium. Governed by d²x/dt² + ω²x = 0.
Angular frequency (ω)
The rate of oscillation in radians per second. Related to frequency and period by ω = 2πf = 2π/T.
Amplitude (A)
The maximum displacement from equilibrium. Determined by initial conditions (initial position x₀ and initial velocity v₀).
Phase constant (φ₀)
Sets where in its cycle the oscillation starts at t = 0. Found from tan φ₀ = −v₀ / (ωx₀).
Period (T)
The time for one complete oscillation. Depends on the system but never on amplitude for ideal SHM.
Frequency (f)
The number of complete oscillations per second. f = 1/T.
The defining equation is:
d²x/dt² + ω²x = 0
This is equivalent to writing a(t) = −ω²x(t), or for angular systems, α(t) = −ω²θ(t).
Any system whose equation of motion takes this form undergoes SHM.
x(t) = A cos(ωt + φ₀)
v(t) = −ωA sin(ωt + φ₀)
a(t) = −ω²A cos(ωt + φ₀)
Key relationships to notice:
Velocity leads position by a quarter cycle (90°). When displacement is maximum, velocity is zero, and vice versa.
Acceleration is always opposite in sign to displacement. Maximum acceleration occurs at maximum displacement.
Peak velocity magnitude: v_max = ωA
Peak acceleration magnitude: a_max = ω²A
Given initial position x₀ and initial velocity v₀:
tan φ₀ = −v₀ / (ωx₀)
A² = x₀² + (v₀/ω)²
Be careful with the quadrant of φ₀ when using arctan. Check that x(0) = x₀ and v(0) = v₀ come out correctly.
Simple pendulum
T = 2π√(L/g)
L is the length of the pendulum, g = 9.8 m/s²
Valid for small angles only
Physical pendulum (compound pendulum)
T = 2π√(I_P / mgd)
I_P is the moment of inertia about the pivot
d is the distance from the pivot to the centre of mass
Reduces to the simple pendulum when the bob is a point mass (I_P = mL², d = L)
Mass-spring system
T = 2π√(m/k)
m is the mass, k is the spring constant
Independent of gravity
Torsional oscillator
T = 2π√(I/κ)
I is the moment of inertia, κ (kappa) is the torsion constant
The angular equivalent of the spring system
ω = 2πf = 2π/T
These hold for all SHM systems. Once you know ω from the system parameters, you have everything.
Summary of the main equations on one view:
Quantity | Formula |
|---|---|
Position | x(t) = A cos(ωt + φ₀) |
Velocity | v(t) = −ωA sin(ωt + φ₀) |
Acceleration | a(t) = −ω²A cos(ωt + φ₀) |
Phase constant | tan φ₀ = −v₀ / (ωx₀) |
Amplitude | A² = x₀² + (v₀/ω)² |
Pendulum period | T = 2π√(L/g) |
Spring period | T = 2π√(m/k) |
Torsion period | T = 2π√(I/κ) |
Physical pendulum | T = 2π√(I_P / mgd) |
⚠️ The phase constant φ₀ trips people up. Remember: tan φ₀ = −v₀ / (ωx₀). The negative sign on v₀ is easy to drop.
⚠️ Amplitude A is always positive. If A² = x₀² + (v₀/ω)² gives you a number, take the positive square root.
⚠️ The period of a simple pendulum depends on length and g, not on mass. The period of a spring system depends on mass and k, not on g.
⚠️ For the physical pendulum formula, I_P is about the pivot, not the centre of mass. You may need the parallel axis theorem: I_P = I_cm + md².
⚠️ Velocity is zero at the extremes of motion (x = ±A). Acceleration is zero at the equilibrium point (x = 0). These are common true/false targets.
⚠️ The exam specifies: do not round intermediate results, and round final answers only to the decimal places stated.
Q: A 0.5 kg mass on a spring (k = 200 N/m) is released from rest at x₀ = 0.1 m. What are ω, T, and A?
A: ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s. T = 2π/20 = 0.3142 s. Since v₀ = 0, A = x₀ = 0.1 m.
Q: For the system above, what is the phase constant φ₀?
A: tan φ₀ = −v₀/(ωx₀) = 0/(20 × 0.1) = 0, so φ₀ = 0. This makes sense: x(0) = A cos(0) = A = 0.1 m.
Q: A simple pendulum has length 2.0 m. What is its period?
A: T = 2π√(L/g) = 2π√(2.0/9.8) = 2π√(0.20408) = 2π(0.45176) = 2.838 s.
Q: At what point in its cycle does a mass on a spring have maximum speed?
A: At the equilibrium position (x = 0), where all the energy is kinetic.
Q: A physical pendulum has I_cm = 0.5 kg m², mass 2 kg, and the pivot is 0.3 m from the centre of mass. What is the period?
A: First find I_P = I_cm + md² = 0.5 + 2(0.3²) = 0.5 + 0.18 = 0.68 kg m². Then T = 2π√(I_P / mgd) = 2π√(0.68 / (2 × 9.8 × 0.3)) = 2π√(0.68/5.88) = 2π√(0.11565) = 2π(0.34007) = 2.137 s.
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