Pc15 - Simple harmonic motion | OCR A-Level Further Mathematics B (MEI)

Simple harmonic motion turns a second-order differential equation into a repeating movement about equilibrium. Pc15 connects the equation to its sine-and-cosine solution, uses initial displacement and velocity to find the constants, and reads amplitude and period from the result. The equation also explains why a larger swing need not take longer.

The equation behind a repeating motion

If an ideal oscillator is pulled twice as far from equilibrium and released, will it take twice as long to complete a cycle? Its displacement changes, but its period does not. The explanation starts with the acceleration law, not a stopwatch.

Take equilibrium as $x=0$ , with positive displacement in a chosen direction. If acceleration is $-4x$ , then at $x=3$ it is $-12$ , while at $x=-3$ it is $12$ . In both cases acceleration points back towards equilibrium, and doubling the displacement doubles its magnitude.

Simple harmonic motion

A motion is simple harmonic motion, or SHM, when its acceleration is proportional to its displacement from a fixed equilibrium position and directed towards that position:

\frac{d^2x}{dt^2}=-\omega^2x,\qquad \omega>0

Here $x$ is displacement, $t$ is time and $\omega$ is angular frequency. A dot denotes differentiation with respect to time: $\dot{x}$ is velocity and $\ddot{x}$ is acceleration. For example, $\ddot{x}=-4x$ is an SHM equation; $\ddot{x}=4x$ is not, because acceleration points away from equilibrium.

The negative sign supplies the restoring direction. The positive number $\omega^2$ determines how quickly the oscillation runs. During one complete cycle, a sine or cosine argument advances by $2\pi$ radians.

SHM solution and period

For $\ddot{x}=-\omega^2x$ , where $\omega>0$ ,

x=p\cos(\omega t)+q\sin(\omega t),\qquad T=\frac{2\pi}{\omega}

Here $p$ and $q$ are constants determined by the initial conditions, and $T$ is the period of a non-stationary motion. If $T$ is known, $\omega=2\pi/T$ .

Not in the formula booklet: learn it.

You may recognise an SHM equation and quote this solution unless the question specifically asks you to solve it by a stated method. The next part shows the auxiliary-equation route as well.

Worked example 1: building a model from a period

Question. An object is modelled as performing SHM about $x=0$ , with period $4$ seconds. Write down its differential equation and general solution.

Plan. Convert the period to angular frequency, square it for the acceleration law, then write the sine-and-cosine solution.

\begin{aligned} \omega&=\frac{2\pi}{4}=\frac\pi2 && \text{one cycle in four seconds} \\ \ddot{x}&=-\frac{\pi^2}{4}x && \text{use }-\omega^2x \\ x&=p\cos\left(\frac{\pi t}{2}\right)+q\sin\left(\frac{\pi t}{2}\right) && \text{general SHM solution} \end{aligned}

Why this step? The coefficient of $x$ is the square of angular frequency, not angular frequency itself.

Answer. $\ddot{x}=-\frac{\pi^2}{4}x$ , with the displayed general solution.

Check. Increasing $t$ by $4$ increases each trigonometric argument by $2\pi$ , returning to the same displacement and velocity.

Worked example 2 changes only the observed period from $4$ to $6$ seconds. A slower oscillation needs a smaller restoring coefficient.

Worked example 2: a longer period

Question. Write down the equation and general solution for SHM of period $6$ seconds about $x=0$ .

Plan. Recalculate $\omega$ and then $\omega^2$ .

\begin{aligned} \omega&=\frac{2\pi}{6}=\frac\pi3 && \text{new angular frequency} \\ \ddot{x}&=-\frac{\pi^2}{9}x && \text{square before substituting} \\ x&=p\cos\left(\frac{\pi t}{3}\right)+q\sin\left(\frac{\pi t}{3}\right) && \text{same solution structure} \end{aligned}

Why this step? A longer period means a smaller $\omega$ , so the acceleration for a fixed displacement has smaller magnitude.

Answer. $\ddot{x}=-\frac{\pi^2}{9}x$ , with the displayed solution.

Check. In $6$ seconds the argument advances by $\frac\pi3(6)=2\pi$ .

The two constants change the particular motion, while the equation's coefficient fixes the period.

Watch out

Wrong: from $\ddot{x}=-9x$ , take $\omega=9$ and $T=2\pi/9$ . Right: $\omega^2=9$ , so $\omega=3$ and $T=2\pi/3$ .

Use a new observed period to build the equation rather than guessing its coefficient.

Using displacement and velocity at the start

Initial displacement determines the cosine coefficient, while initial velocity determines the sine coefficient after differentiation.

You already know

For a constant-coefficient homogeneous equation, the auxiliary equation gives its complementary function. Roots $\pm\omega i$ produce $p\cos(\omega t)+q\sin(\omega t)$ . Differentiate before applying a velocity condition.

For SHM,

\dot{x}=-p\omega\sin(\omega t)+q\omega\cos(\omega t)

At $t=0$ , the sine terms vanish and cosine is $1$ , so $x(0)=p$ and $\dot{x}(0)=q\omega$ . A non-zero starting displacement does not imply the object starts at a turning point: its velocity must also be zero.

Worked example 3: released from rest

Question. Solve $\ddot{x}+4x=0$ , using the auxiliary equation, given $x=3\,\mathrm{cm}$ and $\dot{x}=0$ when $t=0$ . Time is in seconds.

Plan. Obtain the trigonometric general solution, then apply the displacement and differentiated velocity conditions.

\begin{aligned} r^2+4&=0\quad\Longrightarrow\quad r=\pm2i && \text{auxiliary roots} \\ x&=p\cos2t+q\sin2t && \text{general solution} \\ x(0)&=p=3 && \text{initial displacement} \\ \dot{x}&=-2p\sin2t+2q\cos2t && \text{differentiate} \\ \dot{x}(0)&=2q=0\quad\Longrightarrow\quad q=0 && \text{initial velocity} \end{aligned}

Why this step? “Released from rest” gives a velocity condition, so it belongs in $\dot{x}$ , not in $x$ .

Answer. $x=3\cos2t$ centimetres.

Check. $x(0)=3$ , $\dot{x}(0)=0$ , and $\ddot{x}=-12\cos2t=-4x$ .

Worked example 4 changes only the initial velocity, from $0$ to $8\,\mathrm{cm\,s^{-1}}$ . The same initial position is no longer a turning point.

Worked example 4: already moving at the start

Question. Solve $\ddot{x}+4x=0$ for $x(0)=3\,\mathrm{cm}$ and $\dot{x}(0)=8\,\mathrm{cm\,s^{-1}}$ .

Plan. Use the same auxiliary roots and determine the changed sine coefficient from the new velocity.

\begin{aligned} r&=\pm2i\quad\Longrightarrow\quad x=p\cos2t+q\sin2t && \text{same equation} \\ p&=3 && \text{initial displacement} \\ \dot{x}&=-2p\sin2t+2q\cos2t && \text{differentiate first} \\ 2q&=8\quad\Longrightarrow\quad q=4 && \text{apply }\dot{x}(0)=8 \end{aligned}

Why this step? The factor $2$ comes from differentiating $\sin2t$ ; the sine coefficient is half the initial velocity here.

Answer. $x=3\cos2t+4\sin2t$ centimetres.

Check. At $t=0$ , displacement is $3$ and velocity is $8$ . Each term has second derivative $-4$ times itself, so the differential equation is satisfied.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure

Mechanics (Major and Minor)

Statistics (Major and Minor)

Modelling with Algorithms

Numerical Methods

Extra Pure

Further Pure with Technology

Pc15 - Simple harmonic motion

The equation behind a repeating motion

Using displacement and velocity at the start

Continue this lesson.

Learn faster with Nova