Suppose we consider a 1 dimensional medium to be a number of masses connected to each other by springs.

The Shive Wave Machine, seen in this animation, is essentially such a system, the rods in the machine are connected to each other by a wire (you can think of each element as torsional pendulum. When one spine is displaced the next one feels a force pushing in the same direction, if a continuous force is applied to the system the whole system can move to a new equilibrium.

However if a short disturbance is applied a pulse is observed. When the force at the source is removed the rods are driven back to the their equilibrium position by the restoring force, just like in simple harmonic motion. The duration of the pulse on the machine depends on the time the driving force is applied.

We can see that the velocity of the rods is in a different direction and of a different magnitude to the velocity of the pulse.

The displacement in a pulse can be either perpendicular to, or parallel to the direction of motion, resulting in either transverse or longitudinal pulses.

If the force at the source is changed from a transient one to a continuous oscillatory one then the result is a wave. If we freeze the wave at a certain time the displacement of the points can often be represented a sinusoidal function, $D(x)=A\sin\frac{2\pi}{\lambda}x$.

If the wave moves so that it takes a time $T$, the period, for a wavelength $\lambda$ to pass a point we can say that the velocity of a wave is $v=\frac{\lambda}{T}=f\lambda$.

In order to have a wave which at time $t=0$ has a displacement function $D(x)=A\sin\frac{2\pi}{\lambda}x$ propagate with $v$ we can write

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

which using $v=\frac{\lambda}{T}=f\lambda$ can be written

$D(x,t)=A\sin(\frac{2\pi}{\lambda}x-\frac{2\pi t}{T})=A\sin(kx-\omega t)$

We have here defined a new quantity $k$ which is the wave number, $k=\frac{2\pi}{\lambda}$ and expressed the frequency in terms of the angular frequency $\omega=2\pi f$

The velocity of the waves propagation, which we call the phase velocity can be expressed in terms of $\omega$ and $k$

$v=f\lambda=\frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$

The equation for a wave's motion we have just considered

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

is a solution to the a differential equation that we call the wave equation.

$v^2\frac{\partial^{2} D}{\partial x^2}=\frac{\partial^{2} D}{\partial t^2}$

We can check that this is the case by substituting in our equation of motion in to the differential equation

$-v^2(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)=-v^{2}(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)$

We can now look for physical situations that give rise the wave equation, these will define the values of the velocity in terms of the properties of the medium through which the wave propagates.

http://www.animations.physics.unsw.edu.au/waves-sound/travelling-waves/

We define a piece of the string which has length $\Delta x$, we assume that the amplitude of the wave is fairly small so that the angles $\alpha$ and $\beta$ are small. We then set the two forces equal in magnitude $F_{1}=F_{2}=F_{T}$

The vertical force on the element of the string is then

$-F_{T}\sin\beta-F_{T}\sin\alpha$

From Newton's second law

$-F_{T}\sin\beta-F_{T}\sin\alpha=ma_{y}=\mu\Delta x \frac{\partial^{2} D}{\partial t^2}$

where $\mu$ is the linear mass density of the string ($\mu=\frac{m_{string}}{l_{string}}$)

As the angles $\alpha$ and $\beta$ are small we can say that

$\sin\alpha\approx\tan\alpha=\frac{\partial D}{\partial x}|^{x}$ and $\sin\beta\approx\tan\beta=-\frac{\partial D}{\partial x}|^{x+\Delta x}$

We can the rearrange our equation to be $\Large \frac{(\frac{\partial D}{\partial x}|^{x+\Delta x}-\frac{\partial D}{\partial x}|^{x})}{\Delta x}=\frac{\mu}{F_{T}}\frac{\partial^{2} D}{\partial t^2}$

Taking the limit $\Delta x \to 0$ we get $\large \frac{\partial^{2} D}{\partial x^2}=\frac{\mu}{F_{T}}\frac{\partial^{2} D}{\partial t^2}$

Which is the wave equation with $v=\sqrt{\frac{F_{T}}{\mu}}$

Sound can be represented as a longitudinal wave

$D=A\sin(kx-\omega t)$

The change in pressure from the background pressure $P_{0}$ in response to a volume change is related to the bulk modulus $B$

$\Delta P=-B\frac{\Delta V}{V}=-B\frac{A(D_{2}-D_{1})}{A(x_{2}-x_{1})}$ which in the limit of $\Delta x \to 0$ is $\Delta P=-B\frac{\partial D}{\partial x}$

$\Delta P=-BAk\cos(kx-\omega t)$

$D=A\sin(kx-\omega t)$

$\Delta P=-BAk\cos(kx-\omega t)$

$\Delta P$ is the sound pressure level, the deviation from the background pressure.

We now look at the speed of a longitudinal wave, such as sound, from the perspective of the wave equation. The change in pressure from the background pressure $P_{0}$ in response to a volume change is related to the bulk modulus $B$

$\Delta P=-B\frac{\Delta V}{V}=-B\frac{A(D_{2}-D_{1})}{A(x_{2}-x_{1})}$ which in the limit of $\Delta x \to 0$ is $\Delta P=-B\frac{\partial D}{\partial x}$

The force on the element is given by the difference in the pressure

$F=A(P_{1}-P_{2})=A(\Delta P_{1}-\Delta P_{2})$

Newton's Second Law gives us that

$A(\Delta P_{1}-\Delta P_{2})=m\frac{\partial^{2}D}{\partial t^{2}}=\rho A(x_{2}-x_{1})\frac{\partial^{2}D}{\partial t^{2}}$ → $\frac{\Delta P_{2}-\Delta P_{1}}{x_{2}-x_{1}}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

which in the limit $\Delta x \to 0$ is

$\frac{\partial \Delta P}{\partial x}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

and using our result connecting the pressure to the bulk modulus

$-B\frac{\partial^{2}D}{\partial x^{2}}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

which is the wave equation with $v=\sqrt{\frac{B}{\rho}}$

The speed of sound in air depends on temperature according to

$v=(330+0.6T)\mathrm{ms^{-1}}$

where $T$ is temperature in $\mathrm{^{o}C}$.

Waves transport energy. How much?

We can consider each particle in a wave to be undergoing simple harmonic motion with **total** energy given by $E=\frac{1}{2}kA^{2}$ where $k$ is the effective spring constant and $A$ is the amplitude of the wave. The frequency equation for SHM $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ tell us that $k=4\pi^{2}mf^{2}$ and we can rewrite the energy as

$E=2\pi^{2}mf^{2}A^{2}$

The mass in this equation can be written as $m=\rho V$, where $\rho$ is the density of the medium and $V$ is the volume through which the wave passes. If we are looking at 3 dimensional wave that traveling outwards from a point, we can say it travels a distance $l=vt$ in a time interval $t$. This lets us define our volume as $V=Sl$ where S is the cross-sectional area through which the wave passes. Combining all this, $m=\rho V=\rho Sl=\rho Svt$ and the energy transported through a surface $S$ in time $t$ is

$E=2\pi^{2}\rho Svtf^{2}A^{2}$

and the average power

$\bar{P}=\frac{E}{t}=2\pi^{2}\rho Svf^{2}A^{2}$

The intensity is the average power per unit area

$I=\frac{\bar{P}}{S}=2\pi^{2}\rho vf^{2}A^{2}$

As a 3 dimensional wave propagates from a point the area through which the wave passes increases. When the power output is constant, the intensity decreases as $\frac{1}{r^{2}}$, so the ratio of the intensity at $r_{1}$, $I_{1}$, to the intensity at $r_{2}$, $I_{2}$ is

$\frac{I_{2}}{I_{1}}=\frac{\bar{P}/4\pi r_{2}^{2}}{\bar{P}/4\pi r_{1}^{2}}=\frac{r_{1}^2}{r_{2}^{2}}$

In order for the power output to be constant the amplitude must also decrease with $S_{1}A_{1}^2=S_{2}A_{2}^2$ implying $A\propto\frac{1}{r}$

A one dimensional pulse on a string which reaches the end of string will be reflected. The direction of the pulses displacement depends on the boundary condition where the reflection takes place, ie. whether the string is fixed or free.

This can be demonstrated on the Shive wave machine.

When a wave encounters a change in medium there will be some partial reflection with a phase change that depends on whether it is being reflected from a more or less resistive medium .

We can also look at reflection in 2 dimensions, in a virtual ripple tank.

Here we find that the angle of reflection is equal to the angle of incidence.

Here we will only look at refraction and diffraction briefly (we will take a look at them in the ripple tank!), they will be covered in more detail in PHY132when we look at optics.

Refraction occurs when 2 or 3 dimensional waves pass into a medium where they have a different velocity, when they do this they change their direction.

$\frac{\sin\theta_{2}}{\sin\theta_{1}}=\frac{v_{2}}{v_{1}}$

We can see extensive diffraction in the ripple tank around various objects and we can note that the angular spread of waves $\theta$ around an object is

$\theta\approx\frac{\lambda}{l}$

where $\lambda$ is the wavelength of the waves and $l$ is the width of the opening or object about which diffraction occurs.

When more than one wave passes through a region in space at the same time the displacement is the sum of the separate displacements. This may be a vector sum!

When we have two waves traveling in opposite directions on path of fixed length and both ends are fixed

$D_{1}=A\sin(kx-\omega t)$ and $D_{2}=A\sin(kx+\omega t)$

their sum can produce a standing wave

$D=D_{1}+D_{2}=A\sin(kx-\omega t)+A\sin(kx+\omega t)$

Using $\sin\theta_{1}+\sin\theta_{2}=2\sin\frac{1}{2}(\theta_{1}+\theta_{2})\cos\frac{1}{2}(\theta_{1}-\theta_{2})$

$D=2A\sin kx\cos \omega t$

Recall $k=\frac{2\pi}{\lambda}$ and $\omega=2\pi f$

If the boundary condition is that both ends are fixed then $D$ must be zero at $l$ which means that

$kl=\frac{2\pi l}{\lambda}=\pi,2\pi,3\pi,4\pi..$ etc.

or $\lambda=2l,l,2/3l,l/2,..$ etc.

$f=\frac{v}{\lambda}=\frac{v}{2l},\frac{v}{l},\frac{3v}{2l},\frac{2v}{l},..$ etc.

or if we number the harmonics $n=1,2,3,4,..$

$\lambda=\frac{2l}{n}$ and $f=v\frac{n}{2l}$

When we refer to a harmonic, we are describing the frequency as a multiple of the fundamental frequency.

Recall that for waves on a string $v=\sqrt{\frac{F_{T}}{\mu}}$ so if you take a string and stretch it further you need to take in to account both changes in $l$ and $v$.