Recent content by Nerrad
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Double slit experiment with a glass-covered slit of unknown n
So for the first bit you're saying that the thickness of the mirror is equal to 5 times the wavelength of the light?- Nerrad
- Post #3
- Forum: Introductory Physics Homework Help
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Double slit experiment with a glass-covered slit of unknown n
Homework Statement A double-slit experiment uses a helium-neon laser with a wavelength of 633 nm and a slit separation of 12mm. When a thin sheet of glass is placed in front of one of the slits, the interference pattern shifts by 5 fringes. When the experiment is repeated under water, the shift...- Nerrad
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- Diffraction Double slit Double slit experiment Experiment Interference and diffraction Slit
- Replies: 3
- Forum: Introductory Physics Homework Help
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Partial Differential Equation in Special Relativity
Is it simply due to the fact that ##u_{xx}-c^2u_{tt} \neq u_{x'x'}-c^2u_{t't'} ##, so that if ##u## satisfies the wave equation in ##x##,##t## coordinates, then it does not satisfy the same equation in the ##x'##,##t'## coordinates? Or do I deduce it mathematically?- Nerrad
- Post #6
- Forum: Introductory Physics Homework Help
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Partial Differential Equation in Special Relativity
I give up. Can you guide me through this. Please. Thanks- Nerrad
- Post #5
- Forum: Introductory Physics Homework Help
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Partial Differential Equation in Special Relativity
Can you give a little bit more hints than that please? Thanks- Nerrad
- Post #3
- Forum: Introductory Physics Homework Help
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Partial Differential Equation in Special Relativity
Homework Statement (a) Light waves satisfy the wave equation ##u_{tt}-c^2u_{xx}## where ##c## is the speed of light. Consider change of coordinates $$x'=x-Vt$$ $$t'=t$$ where V is a constant. Use the chain rule to show that ##u_x=u_{x'}## and ##u_{tt}=-Vu_{x'}+u_{t'}## Find ##u_{xx},u_{tt},##...- Nerrad
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- Chain rule Differential Differential equation Partial Pdes Relativity Special relativity
- Replies: 6
- Forum: Introductory Physics Homework Help
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Reflection and Transmission of acoustic waves at a boundary
Oh so what you mean is that it doesn't matter what I set the incident amplitude as, because at the end I'm going to get a percentage/ratio of it? Kinda like how for probability it all adds up to 1?- Nerrad
- Post #5
- Forum: Introductory Physics Homework Help
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Reflection and Transmission of acoustic waves at a boundary
Does it have something to do with ratio? To me it's quite vague- Nerrad
- Post #3
- Forum: Introductory Physics Homework Help
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Reflection and Transmission of acoustic waves at a boundary
Homework Statement An interface is formed between a block of aluminium (density = ##2.70 \times 10^3 kg/m^3##, speed of sound =##6.40 \times 10^3m/s##) and a block of copper (density = ##8.96 \times 10^3 kg/m^3##, speed of sound =##4.76 \times 10^3m/s##). Longitudinal waves traveling through...- Nerrad
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- Acoustic Acoustic waves Boundary Reflection Sound waves Transmission Waves
- Replies: 5
- Forum: Introductory Physics Homework Help
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Change of variables in Heat Equation (and Fourier Series)
From the given BCs for ##u##, am I right in saying that BCs for ##v## is ##v(0,t)=v(a,t)=u(0,t)+C=2C##? Also by substituting ##u(x,t)=v(x,t)+C## into the PDE do you mean partially differentiate it then substitute in like $$v=u+C$$ $$...- Nerrad
- Post #3
- Forum: Calculus and Beyond Homework Help
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Change of variables in Heat Equation (and Fourier Series)
Q: Suppose ##u(x,t)## satisfies the heat equation for ##0<x<a## with the usual initial condition ##u(x,0)=f(x)##, and the temperature given to be a non-zero constant C on the surfaces ##x=0## and ##x=a##. We have BCs ##u(0,t) = u(a,t) = C.## Our standard method for finding u doesn't work here...- Nerrad
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- Change Change of variables Fourier Fourier series Heat Heat equation Pde Series Variables
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Rotational kinematics of a spherical rock upon collision
Thanks guys I appreciate this a lot. :)- Nerrad
- Post #22
- Forum: Introductory Physics Homework Help
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Rotational kinematics of a spherical rock upon collision
Then the distance to the centre of mass of the big is constantly changing as the small rock approaches the CoM?- Nerrad
- Post #19
- Forum: Introductory Physics Homework Help
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Rotational kinematics of a spherical rock upon collision
So the angular momentum L = mvr which taking r as a=0.5, gives a result of L=50 which is apparently the right answer. But why would r be a if its rotating about point O?- Nerrad
- Post #17
- Forum: Introductory Physics Homework Help
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Rotational kinematics of a spherical rock upon collision
So the angular momentum of the small rock about point O before the collision is its mass m * distance from point O * its angular velcoity about point O. But the angular velocity is constantly changing?- Nerrad
- Post #15
- Forum: Introductory Physics Homework Help