build up a distribution that's represented by this wave function This is a very difficult partial differential equation to solve so we need to make some further simplifications. The 2-D and 3-D version of the wave equation is, ∂2u ∂t2 = c2∇2u ∂ 2 u ∂ t 2 = c 2 ∇ 2 u… So just what does this do for us? And that means you can look for a solution of the following form: Substituting the preceding equation into the one before it gives you the following: This equation has terms that depend on either. Given any expression of the form \(a\cos x + b\sin x\), it can be rewritten into any one of the following forms: The form you should use may be given to you in a question, but if not, any one will do. At any point we will specify both the initial displacement of the string as well as the initial velocity of the string. This is a straightforward differential equation, and the solution is, Here, C is a constant and k is the wave vector, where, In practice, however, ER is so small that people almost always just ignore, — that is, they assume it to be 1. Because the string has been tightly stretched we can assume that the slope of the displaced string at any point is small. This means that the string will have no resistance to bending. The initial conditions (and yes we meant more than one…) will also be a little different here from what we saw with the heat equation. This leads to. Solution The wave function of the ball can be written where A is the amplitude of the wave function and is its wave number. Our tips from experts and exam survivors will help you through. So, let’s call this displacement \(u\left( {x,t} \right)\). For the wave equation the only boundary condition we are going to consider will be that of prescribed location of the boundaries or. We know that \(\alpha\) is in the first quadrant because \(k\cos \alpha \textgreater0\) and \(k\sin \alpha \textgreater0\). Requiring the wave function to terminate at the right end of the tube gives He’s also been on the faculty of MIT. Solving the Wave Function of R Using the Schrödinger Equation By Steven Holzner If your quantum physics instructor asks you to solve for the wave function of the center of mass of the electron/proton system in a hydrogen atom, you can do so using a modified Schrödinger equation: Finally, we will let \(Q\left( {x,t} \right)\) represent the vertical component per unit mass of any force acting on the string. , it can be rewritten into any one of the following forms: The form you should use may be given to you in a question, but if not, any one will do. This just means, make them equal each other. which is the wave function of the center of mass of the electron/proton system. Trigonometric equations are solved using a double angle formulae and the wave function. is the wave function for a (fictitious) particle of mass m. Steven Holzner is an award-winning author of technical and science books (like Physics For Dummies and Differential Equations For Dummies). These worked examples show the processes you'll need to go through to rewrite an expression in this form. Finally write the equation in the form it was asked for: \[2\sin x^\circ + 5\cos x^\circ = \sqrt {29} \sin (x + 68.2)^\circ\], Write \(\cos 2x - \sqrt 3 \sin 2x\) in the form \(k\cos (2x + \alpha )\) where \(k\textgreater0\) and \(0 \le \alpha \le 2\pi\), \[\cos 2x - \sqrt 3 \sin 2x = k\cos (2x + \alpha )\], \[= k\cos 2x\cos \alpha - k\sin 2x\sin \alpha\], \[= k\cos \alpha \cos 2x - k\sin \alpha \sin 2x\], \(k\cos \alpha\) is the co-efficient of the \(\cos 2x\) term, \(k\sin \alpha\) is the co-efficient of the \(\sin 2x\) term, \[k = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}\]. Calculate \(k\) using the values for the coefficients: \[\tan \alpha ^\circ =\frac{{k\sin \alpha ^\circ }}{{k\cos \alpha ^\circ }} = \frac{5}{2}\], \[\alpha = {\tan ^{ - 1}}\left( {\frac{5}{2}} \right)\], We know that \(\alpha\) is in the first quadrant as \(k\cos \alpha ^\circ \textgreater 0\) and \(k\sin \alpha ^\circ \textgreater 0\). \[\cos 2x - \sqrt 3 \sin 2x = 2\cos \left( {2x + \frac{\pi }{3}} \right)\]. Beyond this interval, the amplitude of the wave function is zero because the ball is confined to the tube. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here. First, we’re now going to assume that the string is perfectly elastic. Because the Schrödinger equation contains terms involving either R or r but not both, the form of this equation indicates that it’s a separable differential equation. Wave equation in 1D (part 1)* • Derivation of the 1D Wave equation – Vibrations of an elastic string • Solution by separation of variables – Three steps to a solution • Several worked examples • Travelling waves – more on this in a later lecture • d’Alembert’s insightful solution to the 1D Wave Equation *Kreysig, 8th Edn, Sections 11.2 – 11.4 .