Simple harmonic motionEdexcel International A Level Further Maths: Revision notes

A particle moves with simple harmonic motion (SHM) about a centre $O$ if its acceleration is always directed towards $O$ and is proportional to its displacement from $O$: $\ddot x=-\omega^2x.$ Here $\omega$ is the angular frequency in rad s$^{-1}$. The minus sign means the acceleration opposes the displacement. The motion is centred on $O$ and oscillates between $x=\pm a$, where $a$ is the amplitude.

To prove SHM in a given situation: choose a variable $x$ for the displacement from the centre (the equilibrium position), use Newton's second law along the line, and show that the equation becomes $\ddot x=-\omega^2x$ for some positive constant $\omega$. Example: a $2$ kg particle with force $18x$ N towards $O$ gives $2\ddot x=-18x$, so $\ddot x=-9x$ and $\omega=3$. The negative sign and the position of $x$ are essential: it must be measured from the centre, where the resultant force is zero.

The following may be quoted without proof: $x=a\cos\omega t\ \text{or}\ x=a\sin\omega t,\qquad v^2=\omega^2(a^2-x^2),\qquad T=\frac{2\pi}{\omega}.$ Use $x=a\cos\omega t$ if $t=0$ at an extreme position and $x=a\sin\omega t$ if $t=0$ at the centre $O$. The speed is greatest at $O$, where $v_{\max}=a\omega$, and zero at $x=\pm a$. The magnitude of the acceleration is greatest at $x=\pm a$, where it equals $a\omega^2$, and zero at $O$. The frequency is $\frac1T=\frac{\omega}{2\pi}$ hertz.

For speed at a given displacement use $v^2=\omega^2(a^2-x^2)$. If two (speed, displacement) pairs are given, substitute both and subtract to find $\omega$ then $a$. For times, solve the trig equation for the smallest positive $t$. A time between two points is the difference of the two times measured from the same origin, provided both are on the same side of the centre and you choose the first occurrences.

A particle moves with SHM: at $x=1.5$ its speed is $4$ and at $x=2$ it is $3$. Then $16=\omega^2(a^2-2.25)$ and $9=\omega^2(a^2-4)$, so $7=1.75\omega^2$, $\omega=2$ and $a^2=6.25$, $a=2.5$. Starting at $O$, $x=2.5\sin2t$. At $x=1.5$: $t=0.322$ s; at $x=2$: $t=0.464$ s; the time between them is $0.142$ s.