2.E: The Classical Wave Equation (Exercises)
- Page ID
- 210791
Solutions to select questions can be found online.
2.1A
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 \[y = A \cos(3x) + B \sin(3x) \nonumber .\]
2.1B
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 \[y = A \cos(10x) + B \sin(10x) \nonumber \]2.1
2.1C
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
\[y = C_1x + C_2\nonumber \]
2.2A
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
\[\boxed{y = \dfrac{12}{13}e^{-7t} + \dfrac{14}{13}e^{6t}}\nonumber \]
2.3A
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: \[T = {2\pi}\sqrt{\dfrac{m}{k}}\nonumber \]
2.3B
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
\[x(t) = A\sin(\omega t) + B\cos(\omega t)\nonumber \]
2.3C
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.
2.3D
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
\[\nu = \dfrac{\omega}{2\pi}\nonumber \]
2.4
Show that the differential equation:
\[\dfrac{d^2y}{dx^2} + y(x) = 0\nonumber \]
has a solution
\[ y(x)= 2\sin x + \cos x \nonumber \]
- Solution
-
\[\dfrac{dy}{dx}= 2 \cos x - \sin x\nonumber \]
\[\dfrac{d^2y}{dx^2}= -2\sin x - \cos x\nonumber \]
so
\[-2 \sin x-\cos x +2 \sin x +\cos x =0\nonumber \]
\[\dfrac{d^2y}{dx^2} + y(x) = 0 \nonumber \]
2.7
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
\[{\left(\dfrac{m}{k}\right)}^{1/2}\pi<t\le {\left(\dfrac{m}{k}\right)}^{1/2}2\pi\nonumber \]
2.11
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 \(v\).
2.13
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.
2.13
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
\[E_{2,2}=(4h^2π^2/8mπ^2a^2)+(4h^2π^2/8mπ^2b^2) = (h^2/2ma^2)+(h^2/2mb^2)\nonumber \]
2.14
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.
2.18
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 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).
- 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.
2.23
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
\[\dfrac{\hbar}{8ma^2} \le \langle{E}\rangle\nonumber \]
2.33
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 \(T = λ/v = 2π/C\).