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2.E: The Classical Wave Equation (Exercises)

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    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. \(\dfrac{d^{2}y}{dx^{2}} + 9y = 0 \)
    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 \[r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\nonumber \] from which we obtain that \(r = \pm 2\). The general solution has form \(Ae^{xr_1} + Be^{xr_2}\). Thus, the solution is \[ y = Ae^{2x}+Be^{-2x}\nonumber. \]
    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 \[y = A \cos(3x) + B \sin(3x) \nonumber .\]


    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. \(\dfrac{d^{2}y}{dx^{2}} + 100y = 0 \)
    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 \[ y = Ae^{i10x}+Be^{-i10x}\nonumber \] or use Euler's formula to find that \[y = A \cos(10x) + B \sin(10x) \nonumber \]2.1


    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. \(\dfrac{d^{2}y}{dx^{2}} = 0 \)

    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

    \[y = C_1x + C_2\nonumber \]


    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 \(y'(0) = 0\)

    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

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


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

    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

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


    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.


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


    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?


    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

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


    Which two functions oscillate with the same frequency?

    1. \(x(t)=\cos( \omega t)\)
    2. \(x(t)=\sin (2 \omega t)\)
    3. \(x(t)=A\cos( \omega t)+B\sin( \omega t)\)
    1. 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 \(v=\dfrac{\omega}{\pi}\).
    2. Therefore, functions A and C oscillate with the same frequency.
    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.


    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:

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


    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

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


    Show that the differential equation:

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

    has a solution

    \[ y(x)= 2\sin x + \cos x \nonumber \]


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

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


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

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


    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.


    We know that

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


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

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

    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

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


    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 \(v\).


    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

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


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


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

    From these expressions we see that


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


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

    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 \(v\).


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


    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.


    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




    \[Ψ(x,y)= X(x)Y(y)\nonumber \]

    and use separation of variables

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

    \[Y \dfrac{\partial^2 X}{\partial x^2} + X \dfrac{\partial^2 Y}{\partial y^2} + \left(\dfrac{8π^2mE }{h^2}\right) XY = 0\nonumber \]

    divide by \(XY\) to get:

    \[\left(\dfrac{1}{X}\right) \dfrac{\partial^2 X}{\partial x^2} + \left(\dfrac{1}{Y}\right) \dfrac{\partial^2 Y}{\partial y^2} + \dfrac{8π^2mE }{h^2} = 0\nonumber \]

    \[\left(\dfrac{1}{X}\right) \dfrac {\partial^2 X}{\partial x^2} + \left(\dfrac{1}{Y}\right) \dfrac{\partial^2 Y}{\partial y^2} = -\dfrac{8π^2mE}{h^2}\nonumber \]


    \[ \left(\dfrac{1}{X}\right) \dfrac{\partial^2 X}{\partial x^2} = -M^2 \nonumber\]


    \[ \left(\dfrac{1}{Y}\right) \dfrac{\partial^2 Y}{\partial y^2} =-N^2\nonumber \]

    so that

    \[\dfrac{8π^2mE }{h^2} =M^2 + N^2\nonumber \]

    Apply the boundary conditions to get:

    \[X(x)=B \sin{Mx} = B \sin \left({\dfrac{n_x \pi x}{a}}\right) \nonumber \]

    for \(n_x =1,2,3...\infty\).


    \[Y(y)=C\sin(Nx) = C\sin \left(\dfrac{n_yπx}{b}\right) \nonumber\]

    for \(n_y=1,2,3...\infty\).


    \[Ψ(x,y)=X(x)Y(y) \nonumber \]


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

    remember that

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

    where \(M=n_xπ/a\) and \(N=n_yπ/b\)


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

    so we get that

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

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

    \[\begin{align*} E_{2,2} &= \dfrac{4h^2π^2}{8mπ^2a^2} +\dfrac{4h^2π^2}{8mπ^2b^2} \\[4pt] &= \dfrac{h^2}{2ma^2} + \dfrac{ h^2}{2mb^2} \end{align*} \]


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

    • 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.


    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).


    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.


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

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


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

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

    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 \(a\).


    From the given information we know that

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


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

    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

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


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


    To prove \(y\) is a wave you can use the wave equation

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

    where \(v\) is the velocity of the wave.

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

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

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

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

    Finally equating these

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


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

    So this is in fact a traveling wave with positive velocity \(v\).

    You can write \(y\) as \(A \cos(Bx - Ct) \)

    where \(B = 2π/λ\) and \(C = 2πv/λ\)

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

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