Wave

A wave is the motion of a field that oscillates at one point and oscillation propagates to neighboring points. It could be shown that the equation of motion of a point in a wave satisfied the wave equation $$\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$$

Wave Properties

The general solution of the wave equation is $$u(x, t) = A\sin(kx - \omega t + \phi)$$ \(A\) is the amplitude of the wave which indicates the highest point can reach from equilibrium. \(\phi\) is the phase constant. It is used to satisfy the initial condition only.

Consider two points on the wave with \(2\pi\) phase difference (1 cycle) at time \(t\). Define their distance \(\lambda\) as the wavelength. \begin{align} (kx' - \omega t + \phi) - (kx - \omega t + \phi) &= 2 \pi \\ k(x' - x) = k\lambda &= 2\pi \\ k &= \frac{2\pi}{\lambda} \end{align} \(k\) is called the wavenumber.

Consider the same point on the wave \(x\) after 1 cycle of time. Define this time interval \(T\) as period and its inverse \(f = \frac{1}{T}\) as frequency. \begin{align} (kx - \omega t' + \phi) - (kx - \omega t + \phi) &= 2\pi \\ \omega T &= 2\pi \\ \omega = \frac{2\pi}{T} &= 2\pi f \end{align} \(\omega\) is called the angular frequency.

Consider a point on the wave at \(x\) at time \(t\) with a particular phase \(\phi'\). After time \(t'\) its neighboring point \(x'\) has the same phase \(\phi'\). \begin{align} (kx' - \omega t' + \phi) - (kx - \omega t + \phi) &= \phi' - \phi' \\ k\Delta x &= \omega \Delta t \\ v_p \equiv \frac{\Delta x}{\Delta t} &= \frac{\omega}{k} = \lambda f \end{align} \(v_p\) is called the phase velocity. It measures the direction and the speed of the phase of the wave propagates.

Wave Packet

If several waves with different wavenumbers interfere each other, a wavepacket is formed. Each with their phase velocity $$v_p = \frac{\omega}{k} $$ The speed of the overall shape propagates is called the group velocity $$v_g \equiv \frac{\partial \omega}{\partial k}$$

Doppler Effect

Let \(c\) be the phase velocity of the wave through the medium. When both the observer and source are stationary $$v = f\lambda$$ If the source of the wave is moving towards to the receiver at a speed \(v_s\), as the number of cycles observed by a stationary receiver per second is the same, the new wavelength will be $$\lambda' = \frac{c-v_s}{f}$$ the wave speed measured by the observer remains the same as there is no relative motion between the observer and the medium, $$c = f'\lambda' = f\lambda$$ The frequency detected will be $$f' = f\frac{c/f}{(c-v_s)/f} = \frac{c}{c-v_s}f$$ If the observer is moving towards from the source of the wave at a speed \(v_r\), by the Galileo transformation, the wave speed measured by the observer is $$v' = c + v_r$$ The frequency detected will be $$f' = \frac{v'}{\lambda} = \frac{c + v_r}{\lambda} = \frac{c + v_r}{c}f$$ Combining the two equations, the frequency would be received due to the motion of the receiver and the source is $$ \bbox[5px,border:2px solid #666] { f' = \frac{c + v_r}{c - v_s}f } $$