Why is the wave equation y = A sin (wt - kx), why phi = - kx? Why not +kx?
472570
Waves are space-time periodic objects.\n\nYour case: A sin(wt - kx) describes a planar wave, where A is the amplitude and φ(x, t) = wt - kx is the phase.\nFor a fixed time t (e.g. t=0) you have a freezed sin wave in space:\n\nA sin(-kx)\n\nA certain point x in space (e.g. x=0) oscillates over time:\n\nA sin(wt)\n\nSince w and k are arbitrary constants you can write the phase:\n\nφ(x, t) = wt - kx \n\nor\n\nφ(x, t) = kx - wt\n\nWhat counts is the movement of wave fronts (points of constant phase):\n\nφ(x, t) = kx - wt = const <=> x =  const/k + wt/k\n\nφ(x, t) = wt - kx = const <=> x = -const/k + wt/k\n\n ==> v = dx/dt = w/k (phase velocity)
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.
609124
Reflection and refraction do not directly prove Huygen's Theorem, at least not in any way I'm familiar with. Diffraction patterns, such as those of single and double slits, are clear examples of Huygen's Theroem in action.\n(We haven't done wave optics yet)
821401
yes, it's a sine wave, and velocity is integral of that, ie a cosine\n(this is considering only fist order effects)
890169
x = r( t - sint); y = r(1 - cost)\ndx/dt = r(1 - cost); dy/dt = rsint\n\ndy/dx = (dy/dt)/(dx/dt)\n= (rsint)/(r(1 - cost))\n= sint/(1 - cost)\n\nSo at the point where t = a\n\nSlope of tangent (m) = sina/(1 - cosa)\nWhen t = a. x = r(a - sina) and y = r(1 - cosa)\nSo tangent is of the form\ny = mx + b\nand it passes through (r(a - sina) , r(1 - cosa))\n\nThus requred tangent is of form \n\ny = sina/(1 - cosa) x + b\nAnd r(1 - cosa) = sina/(1 - cosa) * r(a - sina) + b\nb = r(1 - cosa) (1 - cosa)\n= r[(1 - cosa)² - sina(a - sina)]/(1 - cosa)\n= r[ (1 - 2cosa + cos²a - asina + sin²a)/(1 - cosa)\n= r[(2(1 - cosa) - asina)]/(1 - cosa))\n= 2r - arsina/(1 - cosa))\nSo the equation is\ny = sina/(1 - cosa) x + 2r - arsina/(1 - cosa))\nOR\nxsina - y(1 - cosa)  + 2r(1 - cosa) - rsina = 0\n\nb) Tangent is horizontal when dy/dx = 0\nie sint/(1 - cost) = 0\nwhence sint = 0\nSo t = -π, 0, π (for -π ≤ t ≤ π)\n\nBut when t = 0, cos t = 1 \nSo dy/dx is indeterminate there (0/0)\n\nBy d'Hopitals Rule\nlim sint/(1 - cost) =\nt →0\n\nlim cost/(-sint)  → ±∞\nt →0\n\nSo horizontal tangents when t = -π, π\nie at (-πr, 2r) and (πr, 2r)\n\nThe tangent is vertical when dy/dx is undefined,\nie 1 - cost = 0\nSo cost = 1\nie t = 0 (for -π ≤ t ≤ π) and here the slope becomes undefined as t→0 (see above)\n\nSo vertical tangent at t = 0 ie at (0, 0)
482358
I admire mathematician's concise answers, but in this case some additional explanation may be due:\n\nWhen a wave is partially transmitted and partially reflected the two outgoing waves are shifted in phase from the incident wave. If energy or probability is to be conserved, the difference between the two phase shifts is pi/2 (90°). In this diagram each of the two possible outgoing  beams consist of components from two paths (R=Reflection,T=Transmission):\n\nOutgoing Horizontal: TRR + RRT\nOutgoing Vertical: TRT + RRR\n\nIn the first case the relative phase shift between the two components is 0 and in the second case it's pi (180°) ==> the two components are negatives of each other and interfere destructively.
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
765826
It's because p waves are faster than s waves, so the difference is related to how far away the source is.  The bigger the difference, the farther away.
303175
go to this site:\nhttp://en.wikipedia.org/wiki/Sine_wave
928253
w² = 16w\nSo w² - 16w = 0\nSo w(w - 16) = 0\nSo either w = 0 or w - 16 = 0\nTherefore w = 0, 16
360055
For blue light'\nd Sin th = 2W        W is wave length\nFor a minimum \nd Sin th = (2n+1)W'/2\nSo,  2W=(2n+1)W'/2\nW'=4W/(2n+1)\n  =1440/(2n+1)\nBy putting various values for n , I find that only for the valu n=1, W' is within the vivible range,\nW'=480 nm
437819
Actually, you get X''/X=C and Y''/Y=-C where C does not have to be 0. Also, remember that sums of solutions are solutions, so the solutions you get from separation of variables only give a basis for the vector space of solutions, not all the solutions. The specifics are difficult in this forum, but you have to set up the series and use the boundary conditions to find the coefficients of the series.
277032
I assume you are talking about any wave equation, not just a QM probability density function.\n\nIn any case, what do you mean that you have a "different x for every instant t"?  That is not the case.  I suspect you are thinking of travelling along with the crest or trough of a wave.  As t changes, the location of the crest and trough changes, but that is not the only location on the wave.\n\nWhat the wave equation specifies is the magnitude of the wave at *any* location x, at any time t.  So, if you are at a constant location x', the magnitude of the wave will vary periodically in time at that location. If you were a cork on the water, you would bob up and down at your location as the waves passed you. But similarly, if we were to somehow freeze time, and travel up and down our x axis, the magnitude of the wave would vary periodically in space as well.  So now, rather than a cork bobbing on the water, imagine you're on a jet ski that is going across the water so fast, compared to the speed of the waves, that it's almost like the waves aren't moving.  Yet you bob up and down as you cross the waves that lie in your path in space.\n\nHence, the wave varies both in x and t, independently.
950482
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
49580
NO WAY Why?\n\nF(t) = A Sin(wt)\n\nA called the magnitude : how high the sine function reach\nif A =1 the sine is between 1 and -1\nif A =5 the sine is between 5 and -5\n\nW = 2*pi*f where f is the frequency : how many period during 1 sec\nif w=1 f=1/(2pi) this means at t=2pi we will have one period\nsin(2pi)=sin(0), one period for each 2pi\nif w=2 >>> one period ecah pi\n\n\nSO when the function is expressed by the sum of many sines, many differents frequancies, impossible to regroup them in one frequency\n\nTHe basic frequency called fimdemental and the other frequencies called harmonics (mulpitles of fundemental)\n\nFourrier Series gives the magnitude of all harmonice of unregular from, BUT IT IS NOT POSSIBLE TO JOINT MORE THAN ONE FREQUENCY IN ONE TERM\n\nlike 3x^2+4x^5 since the powers are differents we can't sum or add them in one term
694104
conjugate(sin(z)) = sin(conjugate(z))
567249
because sin x = (e^ix-e^-ix)/2\n\nsin x does not demonstate motion - it is a standing wave.\n\ne^-ix and e^ix, or rather e^-i(x-ct) and e^-i(x+ct) do demonstrate motion with respect to time. for an object that is fixed in space and vibrating, sines and cosines are good, but for a wave propagating through space, exponentials are easier to understand.