May 25, 2021 – Physicspupil

Recall that classical wave in one spatial dimension is described by the wave equation Eq. (1):

$\displaystyle{\frac{\partial^2u}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2u}{\partial t^2}}$ .

(a) Find conditions on the constants $a$ , $f$ , $k$ and $\phi$ so that $u=a\cos (2\pi ft+kx+\phi )$ is a solution of the Eq. (1).

(b) What are the physical meanings of $a$ , $f$ , $k$ and $\phi$ ?

(c) Show that if $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1), then so is $c_1u_1(x,t)+c_2u_2(x,t)$ where $c_1$ , $c_2$ are constants. (Mathematically, we say that the solutions form a linear space or vector space.)

(d) Consequently, $u=\sum_{i=1}^2a_i\cos (2\pi f_it+k_ix+\phi_i)$ is a solution of Eq. (1) provided that each of the two terms is also a solution of Eq. (1). What could be said about the frequency of the wave described by this linear-superpositioned solution?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2265 Modern Physics Homework 1 Solution.)

(a)

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = \frac{\partial}{\partial x}\bigg(\frac{\partial u}{\partial x}\bigg) \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial x}\Big( -ak\sin (2\pi ft+kx+\phi) \Big) \\ & = -ak^2\cos (2\pi ft+kx+\phi) \\ \frac{\partial^2u}{\partial t^2} & = \frac{\partial}{\partial t}\bigg(\frac{\partial u}{\partial t}\bigg) \\ & = \frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial t}\Big( -a(2\pi f)\sin (2\pi ft+kx+\phi) \Big) \\ & = -4a\pi^2f^2\cos (2\pi ft+kx+\phi ) \end{aligned}$

$-ak^2\cos (2\pi ft+kx+\phi )=\displaystyle{-\frac{4a\pi^2f^2}{v^2}\cos (2\pi ft+kx+\phi )}$ ,

we have $k=\pm \displaystyle{\frac{2\pi f}{v}}$ .

There are no constraints on $a$ and $\phi$ .

(b)

$a$ : amplitude
$f$ : frequency
$k$ : wave number
$\phi$ : phase shift of the wave at $x=0$ and $t=0$

(c)

Set $u(x,t)=c_1u_1(x,t)+c_2u_2(x,t)$ .

Condition $\textrm{(i)}:\enspace u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1).

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = c_1\frac{\partial^2u_1}{\partial x^2} + c_2\frac{\partial^2u_2}{\partial x^2} \\ & \stackrel{\textrm{(i)}}{=} \frac{c_1}{v^2}\frac{\partial^2u_1}{\partial t^2} + \frac{c_2}{v^2}\frac{\partial^2u_2}{\partial t^2} \\ & = \frac{1}{v^2}\bigg( c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2} \bigg) \\ \frac{\partial^2u}{\partial t^2} & = c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2}\\ \therefore \enspace & \frac{\partial^2u}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2u}{\partial t^2} \end{aligned}$

That $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1) implies $c_1u_1(x,t)+c_2u_2(x,t)$ is also a solution.

(d) Frequency is not well-defined for linear-superpositioned waves.

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Day: May 25, 2021

202105251602 Homework 1 (Q1)

202105251532 Homework 2 (Q1)