CP 9.7 - Simple harmonic motion

CP 9.7 - Simple harmonic motion

Simple harmonic motion is governed by an acceleration directed towards equilibrium and proportional to displacement. You will solve its differential equation, use initial displacement and velocity, and connect the solution to amplitude, period, speed and turning points.

Acceleration towards equilibrium

If you double an oscillator's starting displacement, does it take twice as long to complete a cycle? In ideal simple harmonic motion, the larger displacement produces a proportionally larger restoring acceleration. The oscillation gets larger, but its period stays the same.

You already know

For displacement x(t)x(t), velocity is v=x˙v=\dot x and acceleration is x¨\ddot x. A dot means differentiation with respect to time. The homogeneous equation x′′+ω2x=0x''+\omega^2x=0 has purely imaginary auxiliary roots and sine/cosine solutions.

Choose the equilibrium position as x=0x=0, with a fixed positive direction. At x>0x>0, the acceleration must be negative; at x<0x<0, it must be positive. A positive proportionality constant alone would accelerate the particle away from equilibrium.

Simple harmonic motion

A particle moves with simple harmonic motion when its acceleration is proportional to its displacement from a fixed equilibrium position and directed towards that position:

x¨=−ω2x,ω>0\ddot x=-\omega^2x,\qquad \omega>0

Here ω\omega is the angular frequency, measured in radians per unit time. The displacement is measured from equilibrium, which need not be the physical origin used in a question.

For example, x¨=−9x\ddot x=-9x has ω=3\omega=3, not 99. The equation x¨=+9x\ddot x=+9x is not simple harmonic motion: its acceleration points away from equilibrium. At equilibrium, acceleration is zero, but velocity can be at its greatest magnitude.

Key idea

The minus sign supplies the restoring direction; the coefficient supplies ω2\omega^2. Zero acceleration at equilibrium does not mean the particle stops there.

Distinguish angular frequency from its square in the next check.

Solving from displacement and velocity

The auxiliary equation r2+ω2=0r^2+\omega^2=0 has roots ±iω\pm\mathrm{i}\omega. Therefore the real displacement solution contains two independent trigonometric terms.

Displacement and velocity in simple harmonic motion

x=Acos⁡ωt+Bsin⁡ωtx=A\cos\omega t+B\sin\omega t v=−Aωsin⁡ωt+Bωcos⁡ωtv=-A\omega\sin\omega t+B\omega\cos\omega t

In particular, x(0)=Ax(0)=A and v(0)=ωBv(0)=\omega B.

Not in the formula booklet: learn it.

The velocity condition fixes BB through a factor of ω\omega. It is not generally equal to BB itself.

Worked example 1: release from rest

Question. Solve x¨=−4x\ddot x=-4x, given x(0)=3x(0)=3, v(0)=0v(0)=0.

Plan. Write the general solution with ω=2\omega=2, differentiate, then apply both conditions.

x=Acos⁡2t+Bsin⁡2tgeneral solutionv=−2Asin⁡2t+2Bcos⁡2tdifferentiateA=3,2B=0initial conditions\begin{aligned} x&=A\cos2t+B\sin2t && \text{general solution}\\ v&=-2A\sin2t+2B\cos2t && \text{differentiate}\\ A&=3,\qquad 2B=0 && \text{initial conditions} \end{aligned}

Why does release from rest remove the sine term? At t=0t=0, the cosine term contributes no velocity, while the sine term contributes 2B2B.

Answer. x=3cos⁡2tx=3\cos2t.

Check. x¨=−12cos⁡2t=−4x\ddot x=-12\cos2t=-4x, and the displacement and velocity conditions both hold.

Worked example 2 changes only the initial velocity. A sine term is now needed to set the starting direction and speed.

Worked example 2: an initial push

Question. Solve x¨=−4x\ddot x=-4x, given x(0)=3x(0)=3, v(0)=4v(0)=4.

Plan. Use the same general solution and replace only the velocity condition.

A=3initial displacement2B=4initial velocityB=2divide by the frequency\begin{aligned} A&=3 && \text{initial displacement}\\ 2B&=4 && \text{initial velocity}\\ B&=2 && \text{divide by the frequency} \end{aligned}

Why include both terms? The particle is not starting at a turning point, so a pure cosine about t=0t=0 cannot fit the data.

Answer. x=3cos⁡2t+2sin⁡2tx=3\cos2t+2\sin2t.

Check. v=−6sin⁡2t+4cos⁡2tv=-6\sin2t+4\cos2t, which gives v(0)=4v(0)=4, and differentiating again gives −4x-4x.

The sign of the initial velocity matters: positive velocity means movement in the chosen positive direction, not movement away from equilibrium in every situation.

Watch out

Wrong: set B=4B=4 in Example 2. Right: use ωB=4\omega B=4, so B=2B=2. Differentiate before applying a velocity condition.

Change both initial numbers in the matched calculation.

Amplitude, period and maximum speed

The coefficients of a mixed sine-and-cosine solution are not its amplitude. Combine them into one shifted cosine to see the displacement's full range.

For x=Acos⁡ωt+Bsin⁡ωtx=A\cos\omega t+B\sin\omega t, write x=Rcos⁡(ωt−δ)x=R\cos(\omega t-\delta). Expansion gives A=Rcos⁡δA=R\cos\delta, B=Rsin⁡δB=R\sin\delta, hence R=A2+B2R=\sqrt{A^2+B^2}. Choose δ\delta in the quadrant determined by both signs, not from a tangent ratio alone.

Motion quantities

For a non-zero SHM solution of amplitude R>0R>0,

T=2πω,∣v∣max⁡=Rω,∣x¨∣max⁡=Rω2T=\frac{2\pi}{\omega},\qquad |v|_{\max}=R\omega,\qquad |\ddot x|_{\max}=R\omega^2

Also,

v2=ω2(R2−x2)v^2=\omega^2(R^2-x^2)

Not in the formula booklet: learn it.

The speed identity follows by squaring x=Rcos⁡(ωt−δ)x=R\cos(\omega t-\delta) and v=−Rωsin⁡(ωt−δ)v=-R\omega\sin(\omega t-\delta), then using sin⁡2+cos⁡2=1\sin^2+\cos^2=1. It gives speed magnitude; the direction must come from the motion or the sign of vv.

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