# 2.E: The Classical Wave Equation (Exercises)

These are homework exercises to accompany Chapter 2 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.

### Q2.1

Find the general solutions to the following differential equations:

1. $$\dfrac{d^{2}y}{dx^{2}} - 4y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 3\dfrac{dy}{dx} - 54y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} + 9y = 0$$

### S1

1. To solve, we realize that the form of the differential equation is that of a quadratic function: $$ar^{2} + br + c = 0$$ where $$a = 1$$, $$b = 0$$, and $$c = - 4$$. We plug these into the quadratic formula [r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] from which we obtain that $$r = \pm 2$$. The general solution has form $$Ae^{xr_1} + Be^{xr_2}$$. Thus, the solution is $y = Ae^{2x}+Be^{-2x}$.
2. In a similar manner to part a, we use the quadratic formula with $$a = 1$$, $$b = -3$$, and $$c = -54$$ and obtain $$y = Ae^{-6x}+Be^{9x}$$.
3. We also use the quadratic formula with $$a = 1$$, $$b =$$, and $$c = 9$$. The difference is that we obtain imaginary roots. The solution to the quadratic equation is $$\pm 3i$$. We can substitute into the general solution and obtain $$y = Ae^{i3x}+Be^{-i3x}$$. Alternatively, we can use Euler's formula to write it as $y = A \cos(3x) + B \sin(3x)$

### Q2.1

Find the general solutions to the following differential equations:

1. $$\dfrac{d^{2}y}{dx^{2}} - 16y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 6\dfrac{dy}{dx} + 27y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} + 100y = 0$$

### S2.1

1. To find the solution we use the characteristic equation: $$ar^{2} + br + c = 0$$ where $$a = 1$$, $$b = 0$$, and $$c = - 4$$. We use the quadratic formula to find that $$r = \pm 4$$.  The general solution to differential equations of this form are $$Ae^{xr_1} + Be^{xr_2}$$. Thus, the solution is $$y = Ae^{4x}+Be^{-4x}$$.
2. Use same steps as part a but with $$a = 1$$, $$b = -6$$, and $$c = 27$$ to find that $$y = Ae^{9x}+Be^{-3x}$$.
3. Similar as above but with $$a = 1$$, $$b =0$$, and $$c = 100$$. We find that we have imaginary roots since $$r = \pm 10$$. We plug back into general solution to get $y = Ae^{i10x}+Be^{-i10x}$ or use Euler's formula to find that $y = A \cos(10x) + B \sin(10x)$

### Q2.1

Find the general solutions to the following differential equations:

1. $$\dfrac{dy}{dx} - 4\sin(x)y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 5\dfrac{dy}{dx}+6y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} = 0$$

### S2.1

a. Begin by moving the $$4\sin(x)y$$ to the right side.

$\dfrac{dy}{dx} = 4\sin(x)y$

Divide both sides by $$y$$ and multiply by $$dx$$

$\dfrac{1}{y}dy = 4\sin(x)dx$

Integrate both sides

$ln(y) = -4\cos(x)+C$

$y = Ce^{-4\cos(x)}$

b. Begin by saying the solution has the form: $$y = e^{rx}$$ where $$r$$ is a constant. Plug in and factor out the $$e^{rx}$$ yields

$e^{rx}(r^2 - 5r + 6) = 0$

Solve for the roots

$$r = 3$$ and $$r = 2$$

$y = Ae^{3x} + Be^{2x}$

c. Simply take the integral twice to yield

$y = C_1x + C_2$

### Q2.2

Practice solving these first and second order homogeneous differential equations with given boundary conditions:

1. $$\dfrac{dy}{dx} = ay$$          with  $$y(0) = 11$$
2. $$\dfrac{d^2y}{dt^2} = ay$$        with $$y(0) = 6$$  and $$y'(0) = 4$$
3. $$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0$$        with  $$y(0) = 2$$ and $$y'(0) = 0$$

### S2.2

a. This first order differential equation can be solved by combining like variables and integrating. Simply multiply both sides by $$\dfrac{dx}{y}$$ to have the form

$\dfrac{dy}{y} = adx$

Integrating both sides,

$\int\dfrac{dy}{y} = \int adx \\ ln|y| = ax +C$

Solving for y we arrive at our general solution,

$y = Ce^{ax}$

U\sing our boundary condition $$y(0) = 11$$ we can solve the particular solution for $$y$$

$y(0) = 11 = Ce^{a(0)} \\ 11 = Ce^0$

Knowing $$e^0 = 1$$ we determine that $$C = 11$$ and we arrive at our final solution

$\boxed{y = 11e^{ax}}$

b. This second order differential equation can be solved by making a typical guess for $$y$$ based on the flavor of the equation, checking accuracy  then solving for the constants $$C_1$$ and $$C_2$$ u\sing the given boundary conditions.

A typical initial guess for this second order differential equation is $$y = e^{\pm\sqrt{a}t}$$. Test this guess by taking the first two derivatives of $$y$$ with respect to $$x$$ and compare to the given problem.

$y_{guess} = e^{\pm\sqrt{a}t}$

Taking the first derivative of $$y_{guess}$$ with respect to $$x$$ we get

$\dfrac{dy_{guess}}{dx} = \pm\sqrt{a}e^{\pm\sqrt{a}t}$

The second derivative of $$y_{guess}$$ with respect to $$x$$ is

$\dfrac{d^2y_{guess}}{dx^2} = ae^{\pm\sqrt{a}t}$

To check how accurate this $$y_{guess}$$ substitute $$\dfrac{d^2y_{guess}}{dx^2}$$ and $$y_{guess}$$ into  $$\dfrac{d^2y}{dt^2} = ay$$.

$ae^{\pm\sqrt{a}t} = ae^{\pm\sqrt{a}t}$

since both sides of the equal sign are the same we know this was a great guess. Then our general solution will have the form

$y(t) = C_1e^{\sqrt{a}t} \ + \ C_2e^{-\sqrt{a}t}$

U\sing our boundary conditions $$y(0) = 6$$ and $$y'(0) = 4$$ we can solve for $$C_1$$ and $$C_2$$

$y(0) = 6 = C_1e^{\sqrt{a}(0)} \ + \ C_2e^{-\sqrt{a}(0)} \\ 6 = C_1 + C_2 \\ \dfrac{dy(0)}{dx} = 4 = \sqrt{a}C_1e^{\sqrt{a}(0)} \ -\sqrt{a}C_2e^{-\sqrt{a}(0)} \\ 4 = \sqrt{a}C_1-\sqrt{a}C_2$

Know we can use the two system of equations to solve for $$C_1$$ and $$C_2$$. Doing the algebra we find that

$C_1 = 3+\dfrac{2}{\sqrt{a}} \\ C_2 = 3-\dfrac{2}{\sqrt{a}}$

Our particular solution then becomes

$\boxed{y(t) = (3+\dfrac{2}{\sqrt{a}})e^{\sqrt{a}t} \ + \ (3-\dfrac{2}{\sqrt{a}})e^{-\sqrt{a}t}}$

c. To solve this second order differential equation we will use a similar method as part b) but we will add an extra step to determine the exponents of a good $$y_{guess}$$. We will determine the exponents or our exponentials by solving for the roots of this equation as if it was a quadratic equation in which

$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0$

becomes

$r^2 + r - 42 = 0$

Solving for the roots we find that

$(r+7)(r-6) \\ r = -7, \ 6$

Knowing this, our exponents to our $$y_{guess}$$ become,

$y_{guess} = e^{-7t} + e^{6t}$

Now we can follow the same process as part b). Take the first and second derivative of our guess:

$\dfrac{dy_{guess}}{dx} = -7e^{-7t}+6e^{6t}$

Second derivative is

$\dfrac{d^2y_{guess}}{dx^2} = 49e^{-7t}+36e^{6t}$

Substituting $$y_{guess}$$ and our first and second derivative into the original differential equation $$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0$$ we find that

$49e^{-7t}+36e^{6t} + -7e^{-7t}+6e^{6t} - 42e^{-7t} - 42e^{6t} = 0 \\ 0=0$

Again, we were able to make a fantastic $$y_{guess}$$. We can then say that our general solution is

$y = C_1e^{-7t} + C_2e^{6t}$

U\sing our boundary conditions $$y(0) = 2$$ and $$y'(0) = 0$$ we can solve for $$C_1$$ and $$C_2$$

$y(0) = 2 = C_1e^{-7(0)} \ + \ C_2e^{6(0)} \\ 2 = C_1 + C_2 \\ \dfrac{dy(0)}{dx} = 0 = -7C_1e^{-7(0)} \ +6C_2e^{6(0)} \\ 0 = -7C_1+6C_2$

Now we can use the two system of equations to solve for $$C_1$$ and $$C_2$$. Doing the algebra we find that

$C_1 = \dfrac{12}{13} \\ C_2 = \dfrac{14}{13}$

Our particular solution then becomes

$\boxed{y = \dfrac{12}{13}e^{-7t} + \dfrac{14}{13}e^{6t}}$

### Q2.3

Prove that $$x(t)$$ = $$\cos(\theta$$) oscillates with a frequency $$\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$

Prove that $$x(t)$$ = $$\cos(\theta$$) also has a period $$T = {2\pi}\sqrt{\dfrac{m}{k}}$$

$$k$$ is the force constant and $$m$$ is mass of the body.

### S2.3

The angular frequency for a harmonic oscillator in units of radian/second is $\omega = \sqrt{\dfrac{k}{m}}$

Angular frequency and frequency are related by: $\omega = 2{\pi}f$

Substitute: $\omega = \sqrt{\dfrac{k}{m}}$ for omega in: $\omega = 2{\pi}f$

So: $\sqrt{\dfrac{k}{m}} = 2{\pi}f$

Solve for f to find the frequency: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

The period T is the inverse of the frequency so: $f = \dfrac{1}{T}$

Substitute the frequency: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

and solve for T: $\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{1}{T}$

So: $T = {2\pi}\sqrt{\dfrac{m}{k}}$

### Q2.3

Try to show that

$x(t)=\sin(\omega t)$

oscillates with a frequency

$\nu = \omega/2\pi$

Explain your reasoning. Can you give another function of x(t) that have the same frequency?

### S2.3

Both the functions of $$\sin(t)$$ and $$cos(t)$$ experience a complete cycle every $$2π$$ radians, so that they will oscillate at a frequency of 1/2π. Thus, the function $$x(t)=\sin(\omega t)$$ will go through $$\omega$$ complete cycles for every $$2π$$ radians, and thus it will have a frequency of

$\nu = \omega/2\pi$

Another example of function $$x(t)$$ that will have the same frequency of $$\frac{ω}{2π}$$ can be

$x(t) = A\sin(\omega t) + B\cos(\omega t)$

### Q2.3

Which two functions oscillate with the same frequency?

1. $$x(t)=\cos( \omega t)$$
2. $$x(t)=\sin (2 \omega t)$$
3. $$x(t)=A\cos( \omega t)+B\sin( \omega t)$$

### S2.3

1. The function goes through $$\omega$$ cycles every 2$$\pi$$ radians so its frequency is $$v=\dfrac{\omega}{2\pi}$$.
2. The 2 shortens the period to $$\pi$$. The function goes through $$\omega$$ cycles every $$\pi$$ radians so its frequency is $$v=\dfrac{\omega}{\pi}$$.
3. Both $$cos( \omega t)$$ and $$\sin( \omega t)$$ have a frequency of $$\dfrac{\omega}{2\pi}$$ so a linear combination of these functions will have the same frequency.

Therefore, functions A and C oscillate with the same frequency.

### Q2.3

Prove that $$x(t)$$ = $$cos(\omega(t)$$) oscillates with a frequency $$v = \dfrac{\omega}{2\pi}$$

Prove that $$x(t)$$ = A $$cos(\omega(t)$$ + B$$\sin(\omega(t)$$ oscillates with the same frequency, $$v = \dfrac{\omega}{2\pi}$$

### S2.3

Angular frequency which is in units of radian/second is $\omega = \dfrac{2}{v\pi}$

thus the frequency is  $v = \dfrac{\omega}{2\pi}$

In terms of $$x(t)$$ = A $$cos(\omega(t)$$ + B$$\sin(\omega(t)$$, both A and B will have the same angular frequency, thus

$v = \dfrac{\omega}{2\pi}$

### Q2.3

Prove x(t)=cos(wt) oscillates with frequency v=w/2π. Also, prove  x(t)=Acos(wt)+B\sin(wt) oscillates with the exact frequency of w/2π.

### S2.3

The cos(t) and \sin(t) functions oscillate with frequency v= 1/2π. This occurs due to the fact that they complete a cycle every 2π radians. In addition, the cos(wt) and \sin(wt) functions also go through a complete w cycle also every 2π radians. Therefore, this indicates that they oscillate with frequency v=w/2π.

It is important to know that a linear combination of such functions such as A cos(wt) + B \sin(wt)  will oscillate with the exact same w/2π frequency.

Note: A cos(wt) + B \sin(wt) is an example used to make it more understandable.

### Q2.4

Show that the differential equation:

$\dfrac{d^2y}{dx^2} + y(x) = 0$

has a solution

$y(x)= \sin x + \cos x$

### S2.4

$\dfrac{dy}{dx}= 2 \cos x - \sin x$

$\dfrac{d^2y}{dx^2}= -2\sin x - \cos x$

so

$-2 \sin x-\cos x +2 \sin x +\cos x =0$

$\dfrac{d^2y}{dx^2} + y(x) = 0$

### Q2.7

For a classical harmonic oscillator, the displacement is given by

${\xi}\left(t\right)=v_0{\left(\dfrac{m}{k}\right)}^{1/2}{\sin \left[{\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t\right]\ }$

Where $$\xi=x-x_0$$

Derive an expression for the velocity as a function of time, and determine the times at which the velocity of the oscillator is zero.

### S2.7

We know that

$\dfrac{dx}{dt}=v$

So

$\dfrac{d\xi}{dt}=v$

$v=v_0{cos \left({\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t\right)\ }$

The times where the velocity is zero is given by setting the inside equal to the zeros of co\sine.

${\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t=\dfrac{2n+1}{2}\pi$

$t={\left(\dfrac{m}{k}\right)}^{1/2}\dfrac{2n+1}{2}\pi$

The acceleration is

$\dfrac{dv}{dt}=a=-v_0{\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}{\sin \left[{\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t\right]\ }$

If the \sin term is positive, the acceleration will be in the negative $$x$$ direction.

\sine is positive from $$0$$ to $${\pi}$$, the acceleration will be in the negative $$x$$ direction for

$0<t\le {\left(\dfrac{m}{k}\right)}^{1/2}\pi$

and in the positive direction from

${\left(\dfrac{m}{k}\right)}^{1/2}\pi<t\le {\left(\dfrac{m}{k}\right)}^{1/2}2\pi$

### Q2.11

Verify that

$Y(x,t) = A \sin \left(\dfrac{2\pi }{\lambda}(x-vt) \right)$

has a frequency $$\nu$$ = $$v$$/$$\lambda$$ and wavelength $$\lambda$$ traveling right with a velocity $$v$$.

### S2.11

All sine and cosine functions oscillate in a wave-like manner, so $$Y(x,t)$$ is a wave. We can write $$Y(x,t)$$ as

$$Y(x,t) = A\sin(Bx - Ct)$$

where

$B = \dfrac{2\pi}{\lambda}$

and

$C = \dfrac{2v\pi}{\lambda}$

From these expressions we see that

Wavelength

$\lambda = \dfrac{2\pi}{B}$

Frequency

$v = \dfrac{C}{2\pi} = \dfrac{v}{\lambda}$

A standing wave has the equation $$y$$$$s$$ = $$A$$\sin($$k$$$$x$$$$s$$ ). The wave equation represented in this problem is related to the equation of a standing wave by $$x$$ = $$x$$$$s$$  + $$v$$$$t$$.  The point $$x$$$$s$$  is arbitrary, and so we set the equation equal to 0. This gives $$x$$ = $$v$$$$t$$, so the wave is traveling right with velocity $$v$$.

### Q2.13

Explain (in words) how to expand the Hamiltonian into two dimensions and use it solve for the energy.

### S2.13

Begin by writing out the the Hamiltonian in two dimensions. This will include partial differential equations with respect to x and y. Then use that and the boundary conditions to use separation of variables to solve for the energy.

### Q2.13

Given that the Schrödinger equation for a two-dimensional box, with sides a and b, is

$\dfrac{∂^2 Ψ}{∂x^2} + \dfrac{∂^2 Ψ}{∂y^2} +\dfrac{(8π^2mE) }{h^2}Ψ(x,y) = 0$

and it has the boundary conditions of

Ψ(0,y)= Ψ (a,y)=0 and Ψ(o,x)= Ψ(x,b)=0

for all x and y values,

show that E2,2=(h2/2ma2)+(h2/2mb2).

### S2.13

Let

$Ψ(x,y)= X(x)Y(y)$

and use separation of variables.

$\dfrac{\partial^2 Ψ(x,y)}{\partial x^2} + \dfrac{\partial^2 Ψ(x,y)}{\partial y^2} +(8π^2mE / h^2) Ψ(x,y) = 0$

$Y\partial^2 X/\partial x^2 +X\partial^2 Y/\partial y^2 +(8π^2mE / h^2) XY = 0$

divide by XY to get:

$(1/X)\partial^2 X/\partial x^2 +(1/Y)\partial^2 Y/\partial y^2 +(8π^2mE / h^2) = 0$

$(1/X)\partial^2 X/\partial x^2 +(1/Y)\partial^2 Y/\partial y^2 =-(8π^2mE / h^2)$

$Let (1/X)\partial^2 X/\partial x^2 =-M^2 and (1/Y)\partial^2 Y/\partial y^2 =-N^2$

so that

$(8π^2mE / h^2) =M^2+N^2$

Apply the boundary conditions to get:

$X(x)=B*\sin{Mx} = B*\sin{\dfrac{n_x \pi x}{a}} for n_x =1,2,3...infty$

and

$Y(y)=C*\sin(Nx) = C*\sin(nyπx/b) for ny=1,2,3...\infty$

since

$Ψ(x,y)=X(x)Y(y)$

then

$$Ψ(x,y)=B*\sin(\dfrac{n_xπx}{a})*C*\sin(\frac{n_yπx}{b})\] remember that $\dfrac{ (8π^2mE}{h2} =M^2+N^2$ where M=nxπ/a and N=nyπ/b\] therefore [(nxπ/a)2 + (nyπ/b)2](8π2mE / h2)=0\] so we get that $E=\left[(\dfrac{n_xπ}{a})^2 + (\dfrac{n_yπ/b})^2\right](h^2/8π^2m)$ now, you can plug in \(n_x=2$$ and $$n_y=2$$ to get

$E2,2=(4h2π2/8mπ2a2)+(4h2π2/8mπ2b2) = (h2/2ma2)+(h2/2mb2)$

### Q2.14

Explain, in words, how to expand the Schrödinger Equations into a 3 dimensional box.

### S2.14

Step 1: Set up the Hamiltonian for 3 dimensions

Step 2: set up the Schrödinger equation

Step 3: use separation of variables to solve it

Step 4: use solutions to solve for the energy.

### Q2.18

Solving for the differential equation for a pendulum gives us the following equation,

$\phi(x)= c_1\cos {\sqrt{\dfrac{g}{L}}} +c_2\sin {\sqrt{\dfrac{g}{L}}}$

Assuming c1=2, c3= 5, g=7 and L=3, what is the position of the pendulum initially? Does this make sense in the real world. Why or why not? (We can ignore units for this problem).

### S2.18

since we are starting at the initial value, we can assume t0=0. Plugging t0=0 into our equation yields.

$\phi(x)= c_1 \cos((0)\sqrt{\dfrac{g}{L}}) +c_2 \sin((0)\sqrt{\dfrac{g}{L}})$

$$\phi (0) = 2$$

In real life, this makes sense because the pendulum has to start with some potential energy (in our case $$\phi= 2$$) so that it can be transferred to kinetic energy and start oscillating with time.

### Q2.23

Consider a Particle of mass m in a one-dimensional box of length a. Its average energy is given by

$\langle{E}\rangle = \dfrac{1}{2m}\langle p^2\rangle$

Because

$\langle{p}\rangle\ = 0$

$\langle p^2\rangle = \sigma^{2}_{p}$

where $$\sigma_p$$ can be called the uncertainty in p. Using the Uncertainty Principle, show that the energy must be at least as large as $$\hbar/8ma^2$$ because $$\sigma_x$$, the uncertainty in x, cannot be larger than a.

### S2.23

From the given information we know that

$\dfrac{\hbar}{2\sigma_p} < \sigma_x \le a$

Then

$\dfrac{\hbar}{2a}\le\sigma_p$

and so

$$\dfrac{\hbar^2}{4a^2}\le\sigma^{2}_{p} \tag{1}$$

We are given that $$\langle{p^2}\rangle = \sigma^{2}_{p}$$ so we write

$$\dfrac{\sigma^{2}_{p}}{2m} \ = \ \dfrac{\langle{p^2}\rangle}{2m} = \langle{E}\rangle \tag{2}$$

Substituting Eq. 1 into Eq. 2 give

$\dfrac{\hbar}{8ma^2} \le \langle{E}\rangle$

### Q2.33

Prove $$y(x, t) = A\cos[2π/λ(x - vt)]$$ is a wave traveling to the right with velocity $$v$$, wavelength $$λ$$, and period $$λ/v$$.

### S2.33

To prove y is a wave you can use the wave equation

2y/∂t2 = V2*2y/∂x2

where V is the velocity of the wave

y/∂t = A2πv/λ*\sin[2π/λ(x - vt)]

2y/∂t2 = -A22v22 * cos[2π/λ(x - vt)]

y/∂x = -A2π/λ*\sin[2π/λ(x - vt)]

2y/∂x2   -A222 * cos[2π/λ(x - vt)]

v2*2y/∂x2  -A222 * cos[2π/λ(x - vt)] * V2

Finally equating these

2y/∂t2 = v2*2y/∂x

gives

-A22v22 * cos[2π/λ(x - vt)]  -A222 * cos[2π/λ(x - vt)] * V2

Leaving only v = V, so this is in fact a traveling wave with positive velocity v.

You can write y as Acos(Bx - Ct)

where B = 2π/λ and C = 2πv/λ

so λ is easily seen to be  2π/B and the period T = λ/v =  2π/C