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Problem statement, equations, and work done:
In quantum mechanics, there is a relation between momentum and wavelength and between energy and frequency. These are:
##p=\hbar k = \frac{h}{\lambda}##
##E = hf = \hbar \omega##
A wave with an amplitude of 10cm is travelling on a string in the +x direction. The distance between wave crests (tops of oscillations) is 0.5 meters and the string oscillates up and down with a period of 0.10 seconds.
[1] Calculate the wavelength, write the equation for the wave and calculate the wave speed.
##T=0.1 s##
##f = 10 Hz##
##\lambda = 0.5 m##
##A = 10 cm##
##k = \frac{2\pi}{\lambda} = 12.566##
##v = \frac{\lambda}{T} = 5 m/s##
##\omega = 2\pi f = 62.8 rad/sec##
##y(x,t) = A sin(kx  \omega t) = 10 sin(12.566x – 62.8t)##
[2] If the tension in the string is 0.01N, determine the mass per unit length of the string.
##F_t = \mu v^2 → \mu = \frac{F_T}{v^2} = \frac{0.01 N}{(5 m/s)^2} = 0.0004 kg/m##
[3] Use the quantum relations above to substitute momentum and energy for k and ω in the wave equation,; make sure to put the constant in the right places.
##p = \hbar k = \frac{h}{\lambda}##
##E = h f = \hbar \omega##
##k = \frac{p}{\hbar} = \frac{h}{\lambda \hbar}##
##\omega = \frac{E}{\hbar} = \frac{h}{f \hbar}##
##Asin(kx\omega t) = A sin (\frac{px}{\hbar}  \frac{Et}{\hbar}) = Asin (\frac{hx}{\lambda \hbar}  \frac{ht}{f \hbar})##
[4] Show that kxωt is invariant under the Lorentz transformation. That is, with the E and p substitutions, show that, for an observer moving in the direction of wave travel, transforming x,t,E and p produces the same expression as in the original frame. It is helpful to think of vectors and dot products here.
OK this is where I am stuck.
[5] Waves carry energy from place to place. To calculate the energy density and power of a wave, start with Power = F•v. Using the note posted to the Canvas Syllabus on the speed of a wave on a string, show that the power is equal to:
##P = F_y \cdot v_y = F_T tan(\theta) \cdot \frac{\partial y}{\partial t}##
Next, use:
##tan(\theta) = \frac{\partial y}{\partial t}##
to get:
##P = F_T \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}##
Given y = Asin(kxωt), calculate the two partial derivatives and write the full expression for the power. Eliminate FT using v2 = FT/μ and substitute for one of the velocity terms the correct combination of wave parameters ω and k.
Finally for the power, calculate the average power during one cycle by calculating the average of cosine squared over one period.
##f_{average} = <f> = \frac{1}{ba} \int_a^b f(x) dx##
##<cos^2 (\theta)> = \frac{1}{2\pi} \int 0^{2\pi} cos^2(\theta) d\theta##
The graph attached may be of help in determining that average value.
[6] Lastly, to get the energy density, use the power as the energy delivered per unit time, so E(Δt) = P Δt and the energy density will be the energy per unit length of string , so Energy density u = E(Δt)/Δx. Carry out these steps to get an expression for the energy density of a wave on a string.
In quantum mechanics, there is a relation between momentum and wavelength and between energy and frequency. These are:
##p=\hbar k = \frac{h}{\lambda}##
##E = hf = \hbar \omega##
A wave with an amplitude of 10cm is travelling on a string in the +x direction. The distance between wave crests (tops of oscillations) is 0.5 meters and the string oscillates up and down with a period of 0.10 seconds.
[1] Calculate the wavelength, write the equation for the wave and calculate the wave speed.
##T=0.1 s##
##f = 10 Hz##
##\lambda = 0.5 m##
##A = 10 cm##
##k = \frac{2\pi}{\lambda} = 12.566##
##v = \frac{\lambda}{T} = 5 m/s##
##\omega = 2\pi f = 62.8 rad/sec##
##y(x,t) = A sin(kx  \omega t) = 10 sin(12.566x – 62.8t)##
[2] If the tension in the string is 0.01N, determine the mass per unit length of the string.
##F_t = \mu v^2 → \mu = \frac{F_T}{v^2} = \frac{0.01 N}{(5 m/s)^2} = 0.0004 kg/m##
[3] Use the quantum relations above to substitute momentum and energy for k and ω in the wave equation,; make sure to put the constant in the right places.
##p = \hbar k = \frac{h}{\lambda}##
##E = h f = \hbar \omega##
##k = \frac{p}{\hbar} = \frac{h}{\lambda \hbar}##
##\omega = \frac{E}{\hbar} = \frac{h}{f \hbar}##
##Asin(kx\omega t) = A sin (\frac{px}{\hbar}  \frac{Et}{\hbar}) = Asin (\frac{hx}{\lambda \hbar}  \frac{ht}{f \hbar})##
[4] Show that kxωt is invariant under the Lorentz transformation. That is, with the E and p substitutions, show that, for an observer moving in the direction of wave travel, transforming x,t,E and p produces the same expression as in the original frame. It is helpful to think of vectors and dot products here.
OK this is where I am stuck.
[5] Waves carry energy from place to place. To calculate the energy density and power of a wave, start with Power = F•v. Using the note posted to the Canvas Syllabus on the speed of a wave on a string, show that the power is equal to:
##P = F_y \cdot v_y = F_T tan(\theta) \cdot \frac{\partial y}{\partial t}##
Next, use:
##tan(\theta) = \frac{\partial y}{\partial t}##
to get:
##P = F_T \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}##
Given y = Asin(kxωt), calculate the two partial derivatives and write the full expression for the power. Eliminate FT using v2 = FT/μ and substitute for one of the velocity terms the correct combination of wave parameters ω and k.
Finally for the power, calculate the average power during one cycle by calculating the average of cosine squared over one period.
##f_{average} = <f> = \frac{1}{ba} \int_a^b f(x) dx##
##<cos^2 (\theta)> = \frac{1}{2\pi} \int 0^{2\pi} cos^2(\theta) d\theta##
The graph attached may be of help in determining that average value.
[6] Lastly, to get the energy density, use the power as the energy delivered per unit time, so E(Δt) = P Δt and the energy density will be the energy per unit length of string , so Energy density u = E(Δt)/Δx. Carry out these steps to get an expression for the energy density of a wave on a string.
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