Simple Harmonic Motion

01 / UNDERSTAND & APPLY

1. SHM is a restoring-acceleration law, not merely repeated motion

Simple harmonic motion occurs when displacement x from a stable equilibrium obeys a = −ω₀²x with constant positive ω₀². Acceleration is proportional to displacement and points toward equilibrium. A motion can be periodic without satisfying this law: a bouncing ball or a large-angle pendulum generally does not have a sinusoidal displacement governed by one constant ω₀.

For a mass m attached to an ideal spring of stiffness k, measuring x from equilibrium gives mẍ = −kx, so ω₀ = √(k/m). The solution is x = A cos(ω₀t + φ), with amplitude A ≥ 0 fixed by initial conditions. Angular frequency is in radians per second; ordinary frequency f = ω₀/(2π) counts cycles per second, and period T = 2π/ω₀.

The phase θ = ω₀t + φ specifies the location within a cycle. The phase constant φ depends on the chosen time origin and whether sine or cosine is used. A different φ does not represent a different natural frequency. If released from the positive extreme at t = 0 with zero velocity, cosine with φ = 0 is convenient; if crossing equilibrium toward positive x, a sine form is often simpler.

The negative restoring sign matters. The equation ẍ = +ω₀²x describes instability with exponential solutions, not SHM. The zero-displacement position must be the actual equilibrium after all constant forces are included. Measuring from the natural length of a vertical spring without accounting for gravity leaves an extra constant term and obscures the simple restoring equation.

02 / UNDERSTAND & APPLY

2. Differentiate to connect displacement, velocity and acceleration

Differentiation gives v = −Aω₀ sinθ and a = −Aω₀² cosθ. Velocity is a quarter cycle out of phase with displacement, while acceleration is opposite in phase to displacement. At either extreme, speed is zero but acceleration magnitude is maximum. At equilibrium, acceleration is zero but speed magnitude is maximum.

Eliminating phase gives v² = ω₀²(A² − x²). This determines speed, not the sign of velocity: the same position is usually crossed twice per cycle in opposite directions. The sign must come from the phase or motion description. Similarly, knowing x alone determines acceleration in ideal SHM but does not determine velocity direction.

Initial values x₀ and v₀ determine amplitude through A² = x₀² + (v₀/ω₀)². With the cosine convention, cosφ = x₀/A and sinφ = −v₀/(Aω₀). Using only inverse cosine misses the sign of velocity and can choose the wrong phase quadrant. The special A = 0 state is rest at equilibrium, with no meaningful oscillation phase.

Uniform circular motion supplies a geometric representation: the projection of a point rotating uniformly around a circle of radius A onto a diameter obeys the same cosine equation. This does not mean the oscillating particle physically travels around the circle. The construction helps measure phase intervals and avoids treating equal displacement intervals as equal time intervals.

03 / UNDERSTAND & APPLY

3. Use phase to calculate travel times and fractions of a cycle

For a known monotonic part of the cycle, solve the displacement equation for the appropriate phase endpoints and use Δt = Δθ/ω₀. Starting at x = +A, the first arrival at x = 0 takes T/4 and the first arrival at −A takes T/2. Starting at equilibrium instead changes which inverse-trigonometric branch is relevant.

Equal distances do not take equal times. The oscillator moves slowly near a turning point and fastest near the centre. From the positive extreme to x = A/2, phase changes from 0 to π/3, giving time T/6. From A/2 to 0 on the same inward trip, phase changes only from π/3 to π/2, giving T/12 despite the same distance A/2.

For an interval visited more than once, add the contributions over the full cycle. The fraction of time spent in a position range is also the probability of finding the ideal oscillator there when observing at a uniformly random phase. It is greater per unit position near the extremes because motion is slower; this is a classical sampling statement, not a quantum probability distribution.

If a question asks when a specified speed first occurs, there may be several positions and several directions compatible with that speed. Use v² to locate possible positions and the initial phase to select the first event. Squaring equations simplifies algebra but discards sign and chronological information that must be restored.

04 / UNDERSTAND & APPLY

4. Energy alternates between motion and displacement

For an ideal mass–spring oscillator, U = kx²/2 and K = mv²/2, with potential zero at equilibrium. Their sum is E = kA²/2 = mω₀²A²/2. At a turning point, all this mechanical energy is potential; at equilibrium it is kinetic. Doubling amplitude quadruples energy while leaving the natural period unchanged within the linear model.

With x = A cosθ, potential energy is E cos²θ and kinetic energy is E sin²θ. Each energy repeats twice per displacement period because squaring removes the sign of displacement or velocity. Equality K = U occurs at |x| = A/√2, not at A/2. Over a full period, the average of each energy is E/2.

For a vertical spring, equilibrium extension is y_eq = mg/k when downward is positive from the natural length. Writing x = y − y_eq gives mẍ = −kx. Gravity shifts the equilibrium but does not change ω₀ = √(k/m) for this ideal model. The effective quadratic potential about equilibrium includes both gravitational and spring energy, differing from kx²/2 only by a constant.

If the support accelerates uniformly, the effective constant body force changes the equilibrium again; a changing support acceleration can drive the oscillator. Do not conclude that gravity can never affect any oscillation frequency: it does affect a pendulum’s restoring torque. Identify which interaction supplies the displacement-dependent restoring force.

05 / UNDERSTAND & APPLY

5. Derive the restoring force for the actual arrangement

Two ideal springs attached in parallel between the same moving mass and a fixed support experience the same extension. Their forces add, giving k_effective = k₁ + k₂. Springs in series carry the same force in a massless-connector quasistatic model; extensions add, giving 1/k_effective = 1/k₁ + 1/k₂. Extra moving masses or distributed spring inertia can invalidate the simple one-coordinate reduction.

A mass between two fixed springs also has restoring stiffness k₁ + k₂ for a small displacement from its equilibrium, provided both springs act linearly and remain appropriately engaged. One extends while the other compresses, but both incremental forces oppose displacement. If a spring can only pull and goes slack, the force law changes during the cycle and the motion need not remain one SHM.

More generally, near a stable equilibrium x₀, expand a smooth potential as U(x) ≈ U(x₀) + U″(x₀)(x − x₀)²/2 when the first derivative vanishes. If U″(x₀) > 0, small motion is approximately harmonic with ω₀² = U″(x₀)/m. This is why SHM occurs in many systems that are not literal springs.

For a rotational coordinate θ, replace mass by the relevant moment of inertia and force by torque. A torsional restoring torque −κθ produces Iθ̈ + κθ = 0 and T = 2π√(I/κ). The stiffness must have the units appropriate to the chosen coordinate. Inserting a torsional stiffness into a translational k/m formula without this distinction is dimensionally inconsistent.

06 / UNDERSTAND & APPLY

6. Pendulum SHM is a small-angle approximation

For a simple pendulum of length ℓ with a point bob and massless inextensible suspension, torque about the pivot is −mgℓ sinθ. With I = mℓ², the exact ideal angular equation is θ̈ + (g/ℓ)sinθ = 0. Replacing sinθ by θ in radians gives SHM for sufficiently small angular amplitude and period T ≈ 2π√(ℓ/g).

The bob mass cancels in this model, but the period depends on length and local effective gravity. Larger amplitude leads to a period longer than the small-angle prediction. The string must remain taut, and the pivot must be fixed in the frame or its acceleration accounted for. A freely falling pendulum apparatus has no ordinary gravitational restoring torque in its cofalling ideal frame.

For a rigid physical pendulum, let d be the distance from pivot to centre of mass and I_p the moment of inertia about the pivot. Small-angle torque −mgdθ gives T = 2π√(I_p/(mgd)) for d > 0. The equivalent simple-pendulum length is I_p/(md). A pivot at the centre of mass has no gravity restoring torque, so the formula cannot be interpreted as a finite pendulum period there.

In an elevator accelerating upward with constant acceleration a, effective downward gravity is g + a and a small pendulum’s period becomes 2π√(ℓ/(g + a)). With downward acceleration a < g, replace it by g − a. If horizontal acceleration is present, the equilibrium direction tilts along the effective gravity vector; use its magnitude for small oscillations about the shifted equilibrium in the ideal setup.

07 / UNDERSTAND & APPLY

7. Damping changes the free motion and removes mechanical energy

For linear viscous damping, the resistive force is −bẋ with b ≥ 0. The free oscillator obeys mẍ + bẋ + kx = 0, with m > 0 and k > 0. Define γ = b/(2m) and ω₀ = √(k/m). For 0 < γ < ω₀, the underdamped solution is x = C e^(−γt)cos(ω_dt + φ), with ω_d = √(ω₀² − γ²). The displacement is not exact constant-amplitude SHM.

The upper and lower envelopes are ±C e^(−γt), but a turning point of x need not occur exactly where the cosine factor equals ±1 because the envelope itself slopes. Adjacent same-sign displacement extrema have a constant ratio e^(−γT_d), where T_d = 2π/ω_d. The logarithmic decrement is ln(|x_n|/|x_(n+1)|) = γT_d.

When γ = ω₀, critical damping has solution (C₁ + C₂t)e^(−ω₀t). When γ > ω₀, two distinct real decaying exponentials replace the sinusoid. These regimes do not execute repeated oscillations. For a standard displacement release from rest, critical damping returns rapidly without overshooting; arbitrary initial conditions require care before making a universal claim about every trajectory’s crossing behavior.

Mechanical energy satisfies dE/dt = −b v² ≤ 0 exactly for the free damped oscillator. In weak damping, a slowly varying cycle-averaged energy decays approximately as e^(−2γt), twice the amplitude-envelope exponent. Instantaneous energy has additional within-cycle variation, so the pure exponential should not be asserted as an exact time-by-time identity for arbitrary damping and phase.

08 / UNDERSTAND & APPLY

8. A drive selects the steady frequency, while resonance selects the size

With sinusoidal forcing, the linear equation is mẍ + bẋ + kx = F₀ cosΩt. The complete motion contains a transient determined by initial conditions plus a steady response at the driving angular frequency Ω. With positive damping the transient eventually decays, leaving a phase-shifted response. The natural frequency and driving frequency are conceptually different even when close numerically.

The steady displacement amplitude is X = F₀/√[(k − mΩ²)² + (bΩ)²]. At low frequency it approaches F₀/k; at high frequency inertia limits the response. For weak damping the displacement peak lies near ω₀. More precisely its nonzero-frequency maximum occurs at Ω = √(ω₀² − 2γ²) when γ < ω₀/√2; strong damping can remove that nonzero-frequency displacement peak.

At Ω = ω₀ with b > 0, displacement lags the drive by π/2 and the amplitude is F₀/(bω₀). Average input power then balances damping loss in steady state. At zero damping and exact resonance there is no finite steady-state amplitude of this form: the driven solution grows with time in the ideal model. Real systems leave the linear model or dissipate energy before an actual infinity occurs.

Resonance is therefore not just “frequency equality makes amplitude infinite.” It is a frequency-dependent response whose size and width depend on damping and on which quantity is measured. Displacement-amplitude resonance, velocity-amplitude response and absorbed-power resonance should not automatically be assigned identical peak frequencies in every formulation.

09 / UNDERSTAND & APPLY

9. Check the equilibrium, coordinate and approximation

Identify the equilibrium and derive the incremental restoring force or torque before selecting a period formula. Then use initial displacement and velocity to choose amplitude and phase. Track signs through a cycle rather than relying on a speed-only equation to decide direction. For timing, convert position boundaries into the correct phase intervals.

Finally decide whether the model is undamped, weakly damped, overdamped or driven. A decaying envelope does not make every instantaneous energy value a simple exponential, and a drive does not force the natural frequency to change to its own frequency. State small-angle, linear-spring and positive-damping assumptions wherever their formulas are used.

• SHM requires acceleration proportional and opposite to displacement from equilibrium.
• Angular frequency differs from cycles per second by 2π.
• Amplitude needs both initial position and velocity.
• Gravity shifts a vertical spring equilibrium but supplies a pendulum’s restoring torque.
• Damped motion has a transient envelope; forced steady motion follows the drive frequency.
• Resonance size and peak location depend on damping and the measured response.