Simple Harmonic Motion
Solving the SHM equation as a second-order homogeneous differential equation, and reading off amplitude and period.
Simple harmonic motion, revisited
A particle undergoes simple harmonic motion (SHM) if its acceleration satisfies $\ddot x=-\omega^2x$, where $x$ is displacement from a fixed centre and $\omega$ is a positive constant. This is a second-order homogeneous differential equation with constant coefficients.
The dots are Newton's notation for differentiating with respect to time: $\dot x=\dfrac{dx}{dt}$ is velocity $v$, and $\ddot x=\dfrac{d^2x}{dt^2}=\dfrac{dv}{dt}$ is acceleration $a$. So $\dot x$ means "one derivative of $x$ with respect to $t$" and $\ddot x$ means "two derivatives of $x$ with respect to $t$" — they're just shorthand for $\dfrac{dx}{dt}$ and $\dfrac{d^2x}{dt^2}$.
Auxiliary equation: $m^2+25=0\Rightarrow m=\pm5i$, so $x=A\cos5t+B\sin5t$.
$\dot x=-5A\sin5t+5B\cos5t$.
At $t=0$: $x=A=4$; $\dot x=5B=0\Rightarrow B=0$.
So $x=4\cos5t$. Here $\omega=5$, giving period $\dfrac{2\pi}5$ and amplitude $4$.
Once the general solution is written as $x=A\cos\omega t+B\sin\omega t$, the constants $A,B$ are fixed by the initial conditions. The period of the motion is always $T=\dfrac{2\pi}\omega$, independent of $A$ and $B$. The amplitude $x_0=\sqrt{A^2+B^2}$; when one of $A,B$ is zero (as is common when displacement or velocity starts at zero), the amplitude is simply the size of the surviving constant.
The maximum speed occurs at the centre of oscillation and equals $\omega x_0$; the maximum acceleration occurs at the extremes and equals $\omega^2x_0$.
Auxiliary equation: $m^2+16=0\Rightarrow m=\pm4i$, so $x=A\cos4t+B\sin4t$.
$\dot x=-4A\sin4t+4B\cos4t$.
At $t=0$: $x=A=0$; $\dot x=4B=-8\Rightarrow B=-2$.
So $x=-2\sin4t$. Here $\omega=4$ and the amplitude is $x_0=2$, so the maximum speed is $\omega x_0=4\times2=8$.
Exam-style questions
Question 1 6 marks
A particle moves such that $\ddot x=-49x$. Given that $x=0$ and $\dot x=14$ when $t=0$, find $x$ in terms of $t$, and state the amplitude of the motion.
$m^2+49=0\Rightarrow m=\pm7i\Rightarrow x=A\cos7t+B\sin7t$. (M1, A1)
$\dot x=-7A\sin7t+7B\cos7t$. At $t=0$: $x=A=0$; $\dot x=7B=14\Rightarrow B=2$. (M1, A1)
$x=2\sin7t$; amplitude $2$. (A1, A1)
Question 2 6 marks
A particle moves such that $\ddot x=-36x$. Given that $x=5$ and $\dot x=0$ when $t=0$, find $x$ in terms of $t$, and find the maximum speed of the particle.
$m^2+36=0\Rightarrow m=\pm6i\Rightarrow x=A\cos6t+B\sin6t$. (M1, A1)
$\dot x=-6A\sin6t+6B\cos6t$. At $t=0$: $x=A=5$; $\dot x=6B=0\Rightarrow B=0$. (M1, A1)
$x=5\cos6t$. Amplitude $x_0=5$, so maximum speed is $\omega x_0=6\times5=30$. (A1, A1)