Thank you for this enlightens lecture and demo, after spending years as an rf technician and receiving a masters degree in eye I finally understand how a cavity filter works
My initial thought as to why 575hz is the strong transient is because QMo at ωo, that the max amplitude of the system is just about 3% below that balanced point. During the start when there is a veritable plethera of frequencies ω, it finds the max amplitude, even though its not at QMo, and in this case just happens to be a resonant frequency of the cavity so it gives a nice, measurable response.
Het is altijd een genot om U aan het werk te zien Walter. Ik ben tbv mijn mede cursisten radio zendamateurs licentie op zoek naar een fragment van uw lezing waar u laat zien wat er gebeurd als een spoel in resonatie komt en de stroom door de spoel etc. u liet dat zien met een lamp meen ik. Het was zo goed uitgelegd dat het perfect is om filters in radios en zenders te begrijpen in jip en janneke taal. Weet helaas dit ( oude) filmpje niet meer te vinden. Weet u nog welke lezing dit was ?
ik denk dat ik een RLC system had waarin ik de L waarde kon wijzigen door de ijzerkern in de spoel in en uit de spoel te bewegen. One can show that, at resonance, XL = XC and thus the impedance (Z) is reduced to R. At resonance, the impedance is the lowest value possible and the current will be the largest value possible. Dan gaat de lamp veel licht geven. Ik herinner me dit allemaal nog maar heel vaag. *Het moet in een van mijn mijn 8.02 lectures zijn nadat ik LRC systemen heb behandeld.* Zoek wat en je zult het vast wel vinden.
Hi Prof.! Sorry for taking up your time, but I have a question regarding this lecture's 15:35-15:58. Since the boundary conditions of the waveguide require a non-zero k_x component, does that mean a k vector without any k_x component and only k_z component (i.e. propagation direction is parallel to the plates) will not pass through and be reflected?
Prof. Lewin, I had a problem about wave guides. Suppose the boundary condition is in the x direction as you demonstrated. For an EM wave that is polarized in the y direction and travels in the xz plane, there would be a cutoff frequency as you have taught in the video. What happens if a wave travels EXACTLY in the z direction (no x-component in k)? I mean, if this wave travels in z direction and its E-field only oscillates along the y-direction, shouldn't it be free from the boundary condition, which is in the x-direction? Following that, is there still a cutoff frequency? Thank you for your time.
It would take me too long to answer your question. I suggest you watch my 8.03 lectures on waveguides. In your problem, observe carefully the boundary conditions (wave guides is ALL about boundary conditions) and use Maxwell's eqs.
Prof. Lewin, thanks for the swift reply. Perhaps I could put my question in another way to check my understanding with you. For a wave that comes in the waveguide at an angle, since the wave has to meet the boundary conditions, it means only if the wave comes in right at a certain angle (to meet the x-direction restrictions) would they be able to propagate through the waveguide. If they come in at the wrong angle, the wave would be attenuated. As the wavelength increases, the angle (with respect to the tangential of the plates) increases. When it reaches the cutoff frequency (correspondingly a maximum wavelength), this angle reaches 90 degrees (i.e. normal to the plates), which means the wave actually cannot propagate. Please kindly guide me if there is anything incorrect in my understanding.
@@lecturesbywalterlewin.they9259 It is really a surprise to hear from you professor. If the wave is reflected, wouldn't it be the violation of law of reflection?? (Angle of incidence is not equal to angle of reflection)
Cleared up all the confusion I still had about phase velocity and wave guides. Thanks.
Thank you for this enlightens lecture and demo, after spending years as an rf technician and receiving a masters degree in eye I finally understand how a cavity filter works
The sound demo in closed box is very interesting, and useful
as I am currently refurbishing my old hi-fi speakers.
Thank you Marcos !
superbly awesome lecture........what a fantastic way of teaching ..god bless him
I really appreciate what you've done.
It's funny, when you're in college you're too busy to come to class, but when you finish college you watch other people's lectures on the internet
My initial thought as to why 575hz is the strong transient is because QMo at ωo, that the max amplitude of the system is just about 3% below that balanced point. During the start when there is a veritable plethera of frequencies ω, it finds the max amplitude, even though its not at QMo, and in this case just happens to be a resonant frequency of the cavity so it gives a nice, measurable response.
Prof. Lewin teach incredibly well and make love the laws of Nature, moreover he is cool, he looks like Clint Eastwood!
Het is altijd een genot om U aan het werk te zien Walter. Ik ben tbv mijn mede cursisten radio zendamateurs licentie op zoek naar een fragment van uw lezing waar u laat zien wat er gebeurd als een spoel in resonatie komt en de stroom door de spoel etc. u liet dat zien met een lamp meen ik. Het was zo goed uitgelegd dat het perfect is om filters in radios en zenders te begrijpen in jip en janneke taal. Weet helaas dit ( oude) filmpje niet meer te vinden. Weet u nog welke lezing dit was ?
ik denk dat ik een RLC system had waarin ik de L waarde kon wijzigen door de ijzerkern in de spoel in en uit de spoel te bewegen. One can show that, at resonance, XL = XC and thus the impedance (Z) is reduced to R. At resonance, the impedance is the lowest value possible and the current will be the largest value possible. Dan gaat de lamp veel licht geven. Ik herinner me dit allemaal nog maar heel vaag. *Het moet in een van mijn mijn 8.02 lectures zijn nadat ik LRC systemen heb behandeld.* Zoek wat en je zult het vast wel vinden.
I hope Ample Mike gets his royalties for his part in this video at 1:10:10. Keep going with the diet Mike.
Marcos and professor Lewin, name a better duo, I'll wait
Hi Prof.! Sorry for taking up your time, but I have a question regarding this lecture's 15:35-15:58. Since the boundary conditions of the waveguide require a non-zero k_x component, does that mean a k vector without any k_x component and only k_z component (i.e. propagation direction is parallel to the plates) will not pass through and be reflected?
Prof. Lewin, I had a problem about wave guides. Suppose the boundary condition is in the x direction as you demonstrated. For an EM wave that is polarized in the y direction and travels in the xz plane, there would be a cutoff frequency as you have taught in the video. What happens if a wave travels EXACTLY in the z direction (no x-component in k)? I mean, if this wave travels in z direction and its E-field only oscillates along the y-direction, shouldn't it be free from the boundary condition, which is in the x-direction? Following that, is there still a cutoff frequency? Thank you for your time.
It would take me too long to answer your question. I suggest you watch my 8.03 lectures on waveguides. In your problem, observe carefully the boundary conditions (wave guides is ALL about boundary conditions) and use Maxwell's eqs.
Prof. Lewin, thanks for the swift reply. Perhaps I could put my question in another way to check my understanding with you.
For a wave that comes in the waveguide at an angle, since the wave has to meet the boundary conditions, it means only if the wave comes in right at a certain angle (to meet the x-direction restrictions) would they be able to propagate through the waveguide. If they come in at the wrong angle, the wave would be attenuated.
As the wavelength increases, the angle (with respect to the tangential of the plates) increases. When it reaches the cutoff frequency (correspondingly a maximum wavelength), this angle reaches 90 degrees (i.e. normal to the plates), which means the wave actually cannot propagate.
Please kindly guide me if there is anything incorrect in my understanding.
if the wave comes in at an angle. decompose it in one direction parallel to the surface and one perpendicular. Use Maxwell's eqs for both separately.
Very smart work
45:37 is that girl Nergis Mavalvala, Astrophysicist, Genius Award winner (Ligo) 2010
Yes she was one of my instructors on 8.03. Instructors do not lecture but they meet meet twice per week with small classes (about 3 students)
16:38 It is said: "This under the square root is simply 'k'".
Wouldn't that be: "Simply k^2".
what I have is correct
At 12:00, why should the cos(wt) not be there? Is it because all the timing information is taken care of in the traveling wave portion?
incorrect
Lectures by Walter Lewin. They will make you ♥ Physics. So the standing part of the wave has no time variance?
question unclear. sin(x)*cos(wt) depends on time
Lectures by Walter Lewin. They will make you ♥ Physics. But you say cos(wt) shouldn't be there
the time term (cos wt) is already in the first term.
I'm getting a 404 error when I try to access any of the assignments or lecture notes in the video descriptions. Is this a problem on my end?
mit.ucu.ac.ug/OcwWeb/Physics/8-03Fall-2004/CourseHome/index.htm
At 34:35, where does the EM wave go if it doesn't pass through??
it's reflected
@@lecturesbywalterlewin.they9259 It is really a surprise to hear from you professor.
If the wave is reflected, wouldn't it be the violation of law of reflection??
(Angle of incidence is not equal to angle of reflection)