Simple harmonic motionEdexcel A-Level Further Maths: Revision notes

A particle moves with simple harmonic motion (SHM) when its acceleration is directed towards a fixed point and proportional to its displacement from it: $\ddot{x}=-\omega^2x.$ The motion is between $x=-a$ and $x=a$, where $a$ is the amplitude. To prove SHM, apply $F=ma$ at a general displacement $x$, show the resultant force is $-kx$, and write it as $\ddot{x}=-\omega^2x$ with $\omega^2=\frac km$. The proof is only valid while the motion stays within the region where the force law holds (for example, strings taut).

The solutions are $x=a\sin\omega t$ (starting at the centre) or $x=a\cos\omega t$ (starting at maximum displacement). These can be quoted without proof, as can: $v^2=\omega^2(a^2-x^2),\quad T=\frac{2\pi}{\omega}.$ The maximum speed is $a\omega$ at the centre; the maximum acceleration is $a\omega^2$ at the ends. Set your calculator to radians when solving for $t$. Example: $a=0.8$, $\omega=5$: $v_{max}=4$ m s⁻¹; at $x=0.4$, $v^2=25(0.64-0.16)=12$, so $v=3.46$ m s⁻¹.

Hooke's law gives the tension in a string or spring: $T=\frac{\lambda e}{l}$, where $\lambda$ is the modulus of elasticity, $l$ the natural length and $e$ the extension. A spring can also be compressed, giving a thrust of the same form. For a particle hanging in equilibrium, $mg=\frac{\lambda e_0}{l}$. At a displacement $x$ below equilibrium, $m\ddot{x}=mg-\frac{\lambda(e_0+x)}{l}=-\frac{\lambda x}{l}$. So the motion is SHM with $\omega^2=\frac{\lambda}{ml}$, and the weight does not affect the period. For a particle between two strings, add the tensions with the correct directions: the resultant is again proportional to $x$.

A string can only pull. If the oscillation takes $P$ above the natural-length position (a distance $e_0$ above equilibrium, so $a>e_0$) the string goes slack and the motion is no longer SHM. Solve in stages: SHM up to the slack point (use $v^2=\omega^2(a^2-x^2)$ with $x=e_0$ above the centre), then free motion under gravity (use $v^2=u^2-2gs$ or energy). Example: $m=0.5$, $l=0.5$, $\lambda=12.25$, released 0.3 m below equilibrium ($e_0=0.2$, $\omega=7$). Slack speed: $v^2=49(0.09-0.04)=2.45$. Further rise $=\frac{2.45}{2(9.8)}=0.125$ m, so the highest point is 0.325 m above equilibrium.

Total mechanical energy is constant. For an oscillation with amplitude $a$: kinetic energy $\frac12mv^2=\frac12m\omega^2(a^2-x^2)$, and the energy at the centre is $\frac12m\omega^2a^2$. For elastic systems include elastic potential energy $\frac{\lambda e^2}{2l}$ and gravitational potential energy $mgh$. Energy gives the speed at a position without finding $\omega$: for the slack example above, $\frac{12.25(0.5)^2}{2(0.5)}=0.5(9.8)(0.5)+\frac12(0.5)v^2$ gives $v^2=2.45$, as before. Kinetic energy is greatest at the centre; the potential energy of the system is greatest at the ends.

To prove SHM: draw the particle at a general displacement $x$ on the positive side, find the resultant force (include every force), apply $F=ma$ and finish with $\ddot{x}=-\omega^2x$. Then quote $T$, $v^2$ or $v_{max}$. Give answers to 3 significant figures and state units. State that the string remains taut when your proof depends on it.