Simple harmonic motionAQA A-Level Further Maths: Revision notes

A particle moves with simple harmonic motion (SHM) about a fixed point $O$ when its acceleration is proportional to its displacement from $O$ and always directed towards $O$: $x''=-\omega^2x,\quad\omega>0.$ The minus sign is the key: it says the force is a restoring force, pulling the particle back towards the equilibrium position $x=0$. The constant $\omega$ is the angular frequency (rad s$^{-1}$). Here $x''$ means $\frac{d^2x}{dt^2}$, the acceleration, and $x'$ is the velocity.

The auxiliary equation of $x''+\omega^2x=0$ is $m^2+\omega^2=0$, so $m=\pm i\omega$ and the general solution is $x=A\cos\omega t+B\sin\omega t.$ This can be written $x=R\cos(\omega t-\alpha)$ with $R=\sqrt{A^2+B^2}$. Then $R$ is the amplitude $a$, the greatest distance from $O$, and the period is $T=\frac{2\pi}{\omega}$, the time for one complete oscillation. The frequency is $f=\frac1T=\frac{\omega}{2\pi}$. Two initial conditions fix $A$ and $B$: use $x(0)$ for $A$, and $x'(0)=B\omega$ for $B$.

Differentiating $x=a\cos(\omega t-\alpha)$ gives $x'=-a\omega\sin(\omega t-\alpha)$. Using $\sin^2+\cos^2=1$: $v^2=\omega^2(a^2-x^2).$ So the speed is greatest at $O$, where $x=0$: $v_{\max}=a\omega$. The speed is zero at the extremes $x=\pm a$ (the particle is momentarily at rest and turns round). The acceleration has magnitude $\omega^2|x|$, so it is greatest at the extremes, $\omega^2a$, and zero at $O$.

For a spring, Hooke's law gives tension $T=kx$, where $x$ is the extension and $k$ is a constant. For a particle of mass $m$ on a smooth horizontal surface attached to a spring, the displacement $x$ from the equilibrium position gives $mx''=-kx$, so $x''=-\frac{k}{m}x$. This is SHM with $\omega^2=\frac{k}{m}$, so the period is $2\pi\sqrt{\frac{m}{k}}$. For a vertical spring, let $y$ be the displacement below equilibrium. At equilibrium $ke=mg$. At displacement $y$ the extension is $e+y$, so $my''=mg-k(e+y)=-ky$. Gravity cancels, so it is still SHM with $\omega^2=\frac{k}{m}$.

A 0.5 kg particle on a smooth horizontal table is attached to a spring with $k=32$ N m$^{-1}$. It is pulled 0.2 m from equilibrium and released from rest. Equation: $0.5x''=-32x$, so $x''=-64x$ and $\omega=8$. Solution: $x=A\cos8t+B\sin8t$. $x(0)=0.2$ gives $A=0.2$, and $x'(0)=0$ gives $B=0$, so $x=0.2\cos8t$. Period: $\frac{2\pi}{8}=\frac{\pi}{4}$ s. Greatest speed: $a\omega=0.2\times8=1.6$ m s$^{-1}$. Greatest acceleration: $\omega^2a=64\times0.2=12.8$ m s$^{-2}$.