4.10f - Simple harmonic motion

4.10f - Simple harmonic motion

Connect a restoring acceleration to a repeating motion. You will solve the simple harmonic motion equation, use displacement and velocity conditions, and interpret amplitude, phase, period and distance travelled in a model.

Restoring acceleration

If you pull an ideal oscillator twice as far from its centre and release it, does its return journey take longer? Its acceleration is also twice as large. For simple harmonic motion, these changes balance exactly: the period does not depend on the amplitude.

Consider a particle of mass 2 kg2\,\mathrm{kg} with displacement x mx\,\mathrm{m} from a fixed equilibrium position. Suppose its resultant force is −18x N-18x\,\mathrm{N}. At x=0.1x=0.1, the force is −1.8 N-1.8\,\mathrm{N}; at x=−0.1x=-0.1, it is +1.8 N+1.8\,\mathrm{N}. Both forces point back towards equilibrium.

You already know

Velocity is v=x˙v=\dot x, acceleration is a=x¨a=\ddot x, and Newton's second law is resultant force =ma=ma. Displacement is signed; speed is ∣v∣|v|.

A negative acceleration is not enough to establish an oscillation: it must be proportional to the signed displacement, with the sign that restores equilibrium.

Simple harmonic motion

A particle performs simple harmonic motion (SHM) when its acceleration is proportional to its displacement from a fixed equilibrium position and directed towards that position. Its equation is x¨=−ω2x\ddot x=-\omega^2x, where the constant ω>0\omega>0 is its angular frequency. For example, x¨=−9x\ddot x=-9x is an SHM equation. The equation x¨=9x\ddot x=9x is not: its acceleration points away from equilibrium.

The zero solution describes a particle remaining at equilibrium. A non-zero solution oscillates. The model assumes that the stated restoring law holds throughout the motion and that no omitted force, such as resistance, changes the resultant.

Worked example 1: a restoring force

Question. A particle of mass 2 kg2\,\mathrm{kg} moves on a straight line with resultant force −18x N-18x\,\mathrm{N}, where x mx\,\mathrm{m} is displacement from equilibrium. Show that the motion is SHM and find ω\omega.

Plan. Use Newton's second law and compare the acceleration with −ω2x-\omega^2x.

2x¨=−18xF=max¨=−9xdivide by the massω2=9⇒ω=3 s−1ω>0\begin{aligned} 2\ddot x&=-18x && F=ma\\ \ddot x&=-9x && \text{divide by the mass}\\ \omega^2=9&\Rightarrow\omega=3\,\mathrm{s^{-1}} && \omega>0 \end{aligned}

Why measure xx from equilibrium? The restoring force then vanishes at x=0x=0, and its sign can be compared directly with the displacement.

Answer. The acceleration is proportional to xx and directed towards equilibrium, so the motion is SHM with ω=3 s−1\omega=3\,\mathrm{s^{-1}}.

Check. At x=0.1x=0.1, the acceleration is −0.9 m s−2-0.9\,\mathrm{m\,s^{-2}}; multiplying by the mass gives the stated force −1.8 N-1.8\,\mathrm{N}.

Change only the mass, keeping the same force law. More mass means less acceleration for the same displacement.

Worked example 2: the effect of mass

Question. The same restoring law F=−18xF=-18x acts on a particle of mass 8 kg8\,\mathrm{kg}. Show that the motion is SHM and find ω\omega.

Plan. Divide the restoring force by the new mass before reading the angular frequency.

8x¨=−18xF=max¨=−94xdivide by 8ω2=94⇒ω=32 s−1positive square root\begin{aligned} 8\ddot x&=-18x && F=ma\\ \ddot x&=-\frac94x && \text{divide by }8\\ \omega^2=\frac94&\Rightarrow\omega=\frac32\,\mathrm{s^{-1}} && \text{positive square root} \end{aligned}

Why take a square root? The coefficient of −x-x is ω2\omega^2, not ω\omega.

Answer. The restoring acceleration gives SHM with ω=32 s−1\omega=\tfrac32\,\mathrm{s^{-1}}.

Check. At x=0.1x=0.1, 8x¨=8(−0.225)=−1.8 N8\ddot x=8(-0.225)=-1.8\,\mathrm{N}, matching the same force law.

For angular displacement θ\theta, an equation such as θ¨=−cθ\ddot\theta=-c\theta, with c>0c>0, has the same mathematical form. If a pendulum model instead gives θ¨=−csin⁡θ\ddot\theta=-c\sin\theta, replacing sin⁡θ\sin\theta by θ\theta is justified only for small angles in radians; the resulting SHM prediction is an approximation.

Watch out

Wrong: x¨=−16x\ddot x=-16x has angular frequency 1616. Right: ω2=16\omega^2=16, so ω=4 s−1\omega=4\,\mathrm{s^{-1}}. A positive sign, x¨=16x\ddot x=16x, would not describe SHM.

Use a new mass to test the same modelling step.

Solving the SHM equation

The differential equation fixes how the motion can vary; the starting displacement and velocity choose its position within that family.

For x¨+ω2x=0\ddot x+\omega^2x=0, the auxiliary equation m2+ω2=0m^2+\omega^2=0 has roots m=±iωm=\pm i\omega. The homogeneous-equation method therefore gives a real sine–cosine solution.

SHM displacement

For constant ω>0\omega>0, the general solution is

x=Acos⁡(ωt)+Bsin⁡(ωt)x=A\cos(\omega t)+B\sin(\omega t)

In the formula booklet.

Differentiating gives the velocity:

v=−Aωsin⁡(ωt)+Bωcos⁡(ωt)v=-A\omega\sin(\omega t)+B\omega\cos(\omega t)

The velocity formula is not in the formula booklet: learn it.

You may quote the displacement form in a modelling question unless asked to derive it or solve the SHM equation. If asked to solve the equation, show the auxiliary equation or an equivalent derivation. Notice the especially useful values x(0)=Ax(0)=A and v(0)=Bωv(0)=B\omega.

Worked example 3: release from rest

Question. Solve x¨=−9x\ddot x=-9x, given x(0)=2x(0)=2 and x˙(0)=0\dot x(0)=0.

Plan. Find the sine–cosine family and apply the displacement condition to xx, and the rest condition to x˙\dot x.

m2+9=0⇒m=±3iauxiliary equationx=Acos⁡3t+Bsin⁡3tgeneral solutionx(0)=A=2initial displacementx˙=−3Asin⁡3t+3Bcos⁡3tdifferentiatex˙(0)=3B=0⇒B=0released from rest\begin{aligned} m^2+9=0&\Rightarrow m=\pm3i && \text{auxiliary equation}\\ x&=A\cos3t+B\sin3t && \text{general solution}\\ x(0)=A&=2 && \text{initial displacement}\\ \dot x&=-3A\sin3t+3B\cos3t && \text{differentiate}\\ \dot x(0)=3B=0&\Rightarrow B=0 && \text{released from rest} \end{aligned}

Why use a separate velocity equation? Being at a particular displacement does not tell you whether the particle is moving forwards, backwards or momentarily at rest.

Answer. x=2cos⁡3tx=2\cos3t.

Check. x¨=−18cos⁡3t=−9x\ddot x=-18\cos3t=-9x, and the two conditions hold at t=0t=0.

Change only the initial velocity, keeping the equation and initial displacement fixed. A sine term is now needed.

Worked example 4: a non-zero initial velocity

Question. Solve x¨=−9x\ddot x=-9x, given x(0)=2x(0)=2 and x˙(0)=6\dot x(0)=6.

Plan. Use the same family and determine the new sine coefficient from the velocity condition.

x=Acos⁡3t+Bsin⁡3tm=±3ix(0)=A=2first conditionx˙=−3Asin⁡3t+3Bcos⁡3tdifferentiatex˙(0)=3B=6⇒B=2second condition\begin{aligned} x&=A\cos3t+B\sin3t && m=\pm3i\\ x(0)=A&=2 && \text{first condition}\\ \dot x&=-3A\sin3t+3B\cos3t && \text{differentiate}\\ \dot x(0)=3B=6&\Rightarrow B=2 && \text{second condition} \end{aligned}

Why does BB equal 22, not 66? Differentiating sin⁡3t\sin3t introduces the factor 33.

Answer. x=2cos⁡3t+2sin⁡3tx=2\cos3t+2\sin3t.

Check. x(0)=2x(0)=2, x˙(0)=6\dot x(0)=6, and each term satisfies x¨=−9x\ddot x=-9x.

A condition at a later time must be substituted at that time; the shortcuts for AA and BB above apply specifically at zero.

Watch out

Wrong: “released from rest” means x(0)=0x(0)=0. Right: it means x˙(0)=0\dot x(0)=0. A particle can start at maximum displacement and still be at rest.

Change the sign of the starting velocity. Its sign tells you which way the particle initially moves.

Amplitude, phase and period

A sum of sine and cosine is one oscillation with a shifted starting point, rather than two different frequencies.

Expanding Rsin⁡(ωt+ϕ)R\sin(\omega t+\phi) gives Rsin⁡ϕcos⁡ωt+Rcos⁡ϕsin⁡ωtR\sin\phi\cos\omega t+R\cos\phi\sin\omega t. Thus A=Rsin⁡ϕA=R\sin\phi and B=Rcos⁡ϕB=R\cos\phi. Squaring and adding gives R=A2+B2R=\sqrt{A^2+B^2} for a non-zero motion.

Amplitude–phase form

The same SHM can be written

x=Rsin⁡(ωt+ϕ),R>0x=R\sin(\omega t+\phi),\qquad R>0

In the formula booklet.

This form makes the size and timing easy to read. For x=Acos⁡ωt+Bsin⁡ωtx=A\cos\omega t+B\sin\omega t, the coefficient comparison and one full turn give the following results.

Amplitude, period and maximum speed

R=A2+B2,T=2πω,vmax⁡=RωR=\sqrt{A^2+B^2},\qquad T=\frac{2\pi}{\omega},\qquad v_{\max}=R\omega

Here RR is the amplitude, TT is the period and vmax⁡v_{\max} is the maximum speed. Choose the phase ϕ\phi to satisfy both Rsin⁡ϕ=AR\sin\phi=A and Rcos⁡ϕ=BR\cos\phi=B.
Not in the formula booklet: learn it.

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