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

- Page ID
- 92352

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

## Q2.1A

Find the general solutions to the following differential equations:

- \(\dfrac{d^{2}y}{dx^{2}} - 4y = 0 \)
- \(\dfrac{d^{2}y}{dx^{2}} - 3\dfrac{dy}{dx} - 54y = 0\)
- \(\dfrac{d^{2}y}{dx^{2}} + 9y = 0 \)

**Solution**-
- 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 \[r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\nonumber \] from which we obtain that \(r = \pm 2\). The general solution has form \(Ae^{xr_1} + Be^{xr_2}\). Thus, the solution is \[ y = Ae^{2x}+Be^{-2x}\nonumber. \]
- In a similar manner to part a, we use the quadratic formula with \(a = 1\), \(b = -3\), and \(c = -54\) and obtain \( y = Ae^{-6x}+Be^{9x}\).
- 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 \[y = A \cos(3x) + B \sin(3x) \nonumber .\]

## Q2.1B

Find the general solutions to the following differential equations:

- \(\dfrac{d^{2}y}{dx^{2}} - 16y = 0 \)
- \(\dfrac{d^{2}y}{dx^{2}} - 6\dfrac{dy}{dx} + 27y = 0\)
- \(\dfrac{d^{2}y}{dx^{2}} + 100y = 0 \)

**Solution**-
- 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 \( y = Ae^{4x}+Be^{-4x}\).
- Use same steps as part a but with \(a = 1\), \(b = -6\), and \(c = 27\) to find that \( y = Ae^{9x}+Be^{-3x}\).
- 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 \[ y = Ae^{i10x}+Be^{-i10x}\nonumber \] or use Euler's formula to find that \[y = A \cos(10x) + B \sin(10x) \nonumber \]Q2.1

## Q2.1C

Find the general solutions to the following differential equations:

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

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

\[\dfrac{dy}{dx} = 4\sin(x)y\nonumber \]

Divide both sides by \(y\) and multiply by \(dx\)

\[\dfrac{1}{y}dy = 4\sin(x)dx\nonumber \]

Integrate both sides

\[ln(y) = -4\cos(x)+C\nonumber \]

\[y = Ce^{-4\cos(x)}\nonumber \]

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

\[e^{rx}(r^2 - 5r + 6) = 0\nonumber \]

Solve for the roots

\(r = 3\) and \(r = 2\)

\[y = Ae^{3x} + Be^{2x}\nonumber \]

c. Simply take the integral twice to yield

\[y = C_1x + C_2\nonumber \]

## Q2.2A

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

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

**Solution**-
**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\[\dfrac{dy}{y} = a\,dx\nonumber \]

Integrating both sides,

\[\int\dfrac{dy}{y} = \int adx \\ \ln|y| = ax +C \nonumber \]

Solving for y we arrive at our general solution,

\[y = Ce^{ax}\nonumber \]

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

\[y(0) = 11 = Ce^{a(0)} \\ 11 = Ce^0\nonumber \]

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

\[\boxed{y = 11e^{ax}}\nonumber \]

**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 \(C_2\) Using 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.

\[y_{guess} = e^{\pm\sqrt{a}t}\nonumber \]

Taking the first derivative of \(y_{guess}\) with respect to \(x\) we get

\[\dfrac{dy_{guess}}{dx} = \pm\sqrt{a}e^{\pm\sqrt{a}t}\nonumber \]

The second derivative of \(y_{guess}\) with respect to \(x\) is

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

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

\[ae^{\pm\sqrt{a}t} = ae^{\pm\sqrt{a}t}\nonumber \]

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

\[y(t) = C_1e^{\sqrt{a}t} \ + \ C_2e^{-\sqrt{a}t}\nonumber \]

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

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

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

\[C_1 = 3+\dfrac{2}{\sqrt{a}} \\ C_2 = 3-\dfrac{2}{\sqrt{a}}\nonumber \]

Our particular solution then becomes

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

**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\[\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0\nonumber \]

becomes

\[r^2 + r - 42 = 0\nonumber \]

Solving for the roots we find that

\[(r+7)(r-6) \\ r = -7, \ 6\nonumber \]

Knowing this, our exponents to our \(y_{guess}\) become,

\[y_{guess} = e^{-7t} + e^{6t}\nonumber \]

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

\[\dfrac{dy_{guess}}{dx} = -7e^{-7t}+6e^{6t}\nonumber \]

Second derivative is

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

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

\[49e^{-7t}+36e^{6t} + -7e^{-7t}+6e^{6t} - 42e^{-7t} - 42e^{6t} = 0 \\ 0=0\nonumber \]

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

\[y = C_1e^{-7t} + C_2e^{6t}\nonumber \]

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

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

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

\[C_1 = \dfrac{12}{13} \\ C_2 = \dfrac{14}{13}\nonumber \]

Our particular solution then becomes

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

## Q2.3A

Prove that \(x(t)\) = \(\cos(\theta\)) oscillates with a frequency

\[\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \nonumber\]

Prove that \(x(t)\) = \(\cos(\theta\)) also has a period

\[T = {2\pi}\sqrt{\dfrac{m}{k}} \nonumber\]

where \(k\) is the force constant and \(m\) is mass of the body.

**Solution**-
The angular frequency for a harmonic oscillator in units of radian/second is \[\omega = \sqrt{\dfrac{k}{m}}\nonumber \]

Angular frequency and frequency are related by: \[\omega = 2{\pi}f\nonumber \]

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

So: \[\sqrt{\dfrac{k}{m}} = 2{\pi}f\nonumber \]

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

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

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

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

So: \[T = {2\pi}\sqrt{\dfrac{m}{k}}\nonumber \]

## Q2.3B

Try to show that

\[x(t)=\sin(\omega t)\nonumber \]

oscillates with a frequency

\[\nu = \omega/2\pi\nonumber \]

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

**Answer**-
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

\[\nu = \omega/2\pi\nonumber \]

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

\[x(t) = A\sin(\omega t) + B\cos(\omega t)\nonumber \]

## Q2.3C

Which two functions oscillate with the same frequency?

- \(x(t)=\cos( \omega t)\)
- \(x(t)=\sin (2 \omega t)\)
- \(x(t)=A\cos( \omega t)+B\sin( \omega t)\)

**Solution**-
- The function goes through \(\omega\) cycles every 2\(\pi\) radians so its frequency is \(v=\dfrac{\omega}{2\pi}\). The 2 shortens the period to \(\pi\). The function goes through \(\omega\) cycles every \(\pi\) radians so its frequency is \(v=\dfrac{\omega}{\pi}\).
- Therefore, functions A and C oscillate with the same frequency.
- 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.

## Q2.3D

Prove that \(x(t) = \cos(\omega(t))\) oscillates with a frequency

\[\nu = \dfrac{\omega}{2\pi} \nonumber.\]

Prove that \(x(t) = A \cos(\omega(t) + B \sin(\omega(t))\) oscillates with the same frequency:

\[\nu = \dfrac{\omega}{2\pi}. \nonumber\]

**Solution**-
Angular frequency which is in units of radian/second is

\[\omega = \dfrac{2}{v\pi}\nonumber \]

thus the frequency is \[\nu = \dfrac{\omega}{2\pi}\nonumber \]

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

\[\nu = \dfrac{\omega}{2\pi}\nonumber \]

## Q2.4

Show that the differential equation:

\[\dfrac{d^2y}{dx^2} + y(x) = 0\nonumber \]

has a solution

\[ y(x)= \sin x + \cos x \nonumber \]

**Solution**-
\[\dfrac{dy}{dx}= 2 \cos x - \sin x\nonumber \]

\[\dfrac{d^2y}{dx^2}= -2\sin x - \cos x\nonumber \]

so

\[-2 \sin x-\cos x +2 \sin x +\cos x =0\nonumber \]

\[\dfrac{d^2y}{dx^2} + y(x) = 0 \nonumber \]

## Q2.7

For a classical harmonic oscillator, the displacement is given by

\[ \xi (t)=v_0 \sqrt{\dfrac{m}{k}} \sin \sqrt{\dfrac{k}{m}} t\ \nonumber \]

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.

**Solution**-
We know that

\[\dfrac{dx}{dt}=v\nonumber \]

So

\[\dfrac{d\xi}{dt}=v\nonumber \]

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

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

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

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

The acceleration is

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

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

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

and in the positive direction from

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

### Q2.11

Verify that

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

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

**Solution**-
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

\[Y(x,t) = A\sin(Bx - Ct)\nonumber \]

where

\[B = \dfrac{2\pi}{\lambda}\nonumber \]

and

\[C = \dfrac{2v\pi}{\lambda}\nonumber \]

From these expressions we see that

Wavelength

\[\lambda = \dfrac{2\pi}{B}\nonumber \]

Frequency

\[v = \dfrac{C}{2\pi} = \dfrac{v}{\lambda}\nonumber \]

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.

**Solution**-
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

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

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

\[E_{2,2}=\left(\dfrac{h^2}{2ma^2}\right)+\left(\dfrac{h^2}{2mb^2}\right).\]

**Solution**-
Let

\[Ψ(x,y)= X(x)Y(y)\nonumber \]

and use separation of variables.

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

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

divide by XY to get:

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

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

Let

\[(1/X)\partial^2 X/\partial x^2 =-M^2 and (1/Y)\partial^2 Y/\partial y^2 =-N^2\nonumber \]

so that

\[8π^2mE / h^2) =M^2+N^2\nonumber \]

Apply the boundary conditions to get:

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

and

\[Y(y)=C\sin(Nx) = C\sin(n_yπx/b) for ny=1,2,3...\infty\nonumber \]

since

\[Ψ(x,y)=X(x)Y(y) \nonumber \]

then

\(Ψ(x,y)=B \sin(\dfrac{n_xπx}{a}) C \sin(\frac{n_yπx}{b})\nonumber \]

remember that

\[\dfrac{ 8\pi^2mE}{h2} =M^2+N^2\nonumber \]

where \(M=n_xπ/a\) and \(N=n_yπ/b\)

therefore

\[\left(\dfrac{n_xπ}{a}\right)^2 + \left(\dfrac{n_yπ}{b}\right)^2](8π^2mE / h^2)=0 \nonumber \]

so we get that

\[E=\left[(\dfrac{n_xπ}{a})^2 + (\dfrac{n_yπ/b})^2\right](h^2/8\pi^2m)\nonumber \]

now, you can plug in \(n_x=2\) and \(n_y=2\) to get

\[E_{2,2}=(4h^2π^2/8mπ^2a^2)+(4h^2π^2/8mπ^2b^2) = (h^2/2ma^2)+(h^2/2mb^2)\nonumber \]

### Q2.14

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

**Solution**-
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,

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

Assuming c_{1}=2, c_{3}= 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).

**Solution**-
since we are starting at the initial value, we can assume \(t_0=0\). Plugging \(t_0=0\) into our equation yields.

\[ \phi(x)= c_1 \cos \left((0)\sqrt{\dfrac{g}{L}}\right) +c_2 \sin \left((0)\sqrt{\dfrac{g}{L}}\right) \nonumber \]

\( \phi (0) = 2 \)

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

\[\langle{E}\rangle = \dfrac{1}{2m}\langle p^2\rangle\nonumber \]

Because

\[\langle{p}\rangle\ = 0\nonumber \]

\[\langle p^2\rangle = \sigma^{2}_{p}\nonumber \]

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 \(a\).

**Solution**-
From the given information we know that

\[\dfrac{\hbar}{2\sigma_p} < \sigma_x \le a\nonumber \]

Then

\[\dfrac{\hbar}{2a}\le\sigma_p\nonumber \]

and so

\(\dfrac{\hbar^2}{4a^2}\le\sigma^{2}_{p} \label{1}\)

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

\(\dfrac{\sigma^{2}_{p}}{2m} \ = \ \dfrac{\langle{p^2}\rangle}{2m} = \langle{E}\rangle \label{2}\)

Substituting Equation \ref{1} into Equation \ref{2} give

\[\dfrac{\hbar}{8ma^2} \le \langle{E}\rangle\nonumber \]

## 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 \(λ/v\).

**Answer**-
To prove y is a wave you can use the wave equation

\[\dfrac{ ∂^2y}{∂t^2} = V^2*∂^2y/∂x^2\nonumber \]

where V is the velocity of the wave

\[ \dfrac{∂y}{∂} = A^2πv/λ*\sin[2π/λ(x - vt)z\nonumber \]

\[ \dfrac{∂^2y}{∂t^2} = -A^24π^2v^2/λ^2 \cos[2π/λ(x - vt)]\nonumber \]

\[ \dfrac{∂y}{∂x} = -A^2π/λ \sin[2π/λ(x - vt)]\nonumber \]

\[ \dfrac{∂^2y}{∂x^2} = -A^24π^2/λ^2 \cos[2π/λ(x - vt)]\nonumber \]

\[ v^2 \dfrac{∂^2y}{∂x^2} = -A^24π^2/λ^2 \cos[2π/λ(x - vt)] V^2\nonumber \]

Finally equating these

\[ \dfrac{∂^2y}{∂t^2} = v^2 \dfrac{∂^2y}{∂x} \nonumber \]

gives

\[ -\dfrac{A^24π^2v^2}{λ^2} \cos[2π/λ(x - vt)] = -A^24π^2/λ^2 \cos[2π/λ(x - vt)] V^2\nonumber \]

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 \(T = λ/v = 2π/C\).