Simple Harmonic Motion (Edexcel A Level Further Maths: Core Pure): Revision Note

Simple harmonic motion (SHM) equations

What is simple harmonic motion?

• A particle undergoing simple harmonic motion moves along a straight line subject to the following constraints:

  • The acceleration of the particle is always directed towards a fixed point on the line of motion

  • The acceleration is proportional to the displacement of the particle from the fixed point

• As a result the particle oscillates back and forth along the line around the fixed point

• Many physical systems can be modelled using simple harmonic motion

  • One example is an object attached to a spring oscillating in one dimension, when friction, air resistance and other such resistive forces are disregarded

What is the equation that describes simple harmonic motion?

• The standard form of the simple harmonic motion equation is $\ddot{x}=-{\omega }^{2}x$

  • x is the displacement of the particle from the fixed point

    • The fixed point is normally indicated by O and is the origin (i.e. zero point) of the coordinate system

    • The fixed point O is known as the centre of oscillation

  • ${\omega }^2$ is the constant of proportionality, and represents the strength of the force accelerating the particle back towards point O

    • The negative sign means that the acceleration is always directed back towards O

    • We use ${\omega }^2$ to assure that the constant is positive, and also to simplify the notation for the solution to the equation

  • $\ddot{x}=\frac{d^{2}x}{d{t}^2}$ is the acceleration of the particle

    • With simple harmonic motion, Newton’s ‘dot notation’ is often used for the derivatives

    • In this notation, $\ddot{x}=\frac{d^{2}x}{d{t}^2}$ and $\stackrel{.}{x}=\frac{dx}{dt}$

Solutions to SHM equations

What is the solution to the standard simple harmonic motion equation?

• The SHM equation may be solved using the standard techniques for second order differential equations

  • $\ddot{x}=-{\omega }^{2}x$ may be rearranged to give the homogeneous second order equation $\ddot{x}+{\omega }^{2}x=0$

  • That has auxiliary equation  $m^2+{\omega }^2=0$ with roots $m=\pm \omega \mathrm{i}$

  • Therefore the general solution is $x=A\cos \omega t+B\sin \omega t$

    • Initial or boundary conditions given with a question may allow you to find the precise values of the arbitrary constants A and B

• The linear combination of sine and cosine terms in the general solution may be rewritten in the form $x=R\sin \left(\omega t+\alpha \right)$

  • This can sometimes be a more useful form for the general solution

  • This form of the equation shows that the particle oscillates around the fixed point O (i.e., around the centre of oscillation)

  • The amplitude of the motion is R

    • $R=\sqrt{A^2+B^2}$

    • This is the maximum distance that the particle moves away from the fixed point O

  • The period of the motion is $\frac{2\mathrm{\pi }}{\omega }$

    • This is the length of time it takes the particle to complete one oscillation

  • The constant α is a ‘phase constant’

    • $\alpha ={\tan }^{-1}\left(\frac{A}{B}\right)$ 

    • If α = 0, then the particle is at point O (i.e., x = 0) when t = 0

    • If $\alpha =\frac{\mathrm{\pi }}{2}$ then the particle is at its maximum displacement (i.e., x = R) when t = 0

How can I solve the simple harmonic motion equation to link displacement x and velocity v?

• Because acceleration is the derivative of velocity, and velocity is the derivative of displacement, we may use the chain rule to write

$\ddot{x}=\frac{dv}{dt}=\frac{dx}{dt}\times \frac{dv}{dx}=v\frac{dv}{dx}$

• Substituting that into the standard SHM equation gives

$v\frac{dv}{dx}=-{\omega }^{2}x$

• That form of the SHM equation may be solved using separation of variables:

  $\int vdv=\int -{\omega }^{2}xdx\Rightarrow \frac{1}{2}{v}^2=-\frac{1}{2}{\omega }^2{x}^2+c$

  • But the velocity is momentarily zero when the particle reaches its maximum displacement

    • After this the particle’s velocity changes direction and the particle heads back towards the centre of oscillation

  • This gives us a boundary condition (because we know that v = 0 when x = R) and allows us to find the value of the constant of integration:

    $\frac{1}{2}{\left(0\right)}^2=-\frac{1}{2}{\omega }^2{\left(R\right)}^2+c\Rightarrow c=\frac{1}{2}{\omega }^2{R}^2$

  • Substituting that value of c into the solution and rearranging gives

    $v^2={\omega }^2\left(R^2-x^2\right)$

    • where R again is the amplitude of the simple harmonic motion

• Note that this version of the solution connects the velocity v to the displacement x, and is independent of the time t

  • Taking square roots gives $v=\pm \omega \sqrt{R^2-x^2}$

  • Be careful when using this to answer questions

    • The direction of the velocity (plus or minus) will depend on the displacement (plus or minus) and whether the particle is moving towards or away from the centre of oscillation