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Let’s consider a point $$x$$ on the string in its equilibrium position, i.e. Our tips from experts and exam survivors will help you through. Here we have a 2nd order time derivative and so we’ll also need two initial conditions. but not both. He graduated from MIT and did his PhD in physics at Cornell University, where he was on the teaching faculty for 10 years. Practice and Assignment problems are not yet written. This means that the string will have no resistance to bending. 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. In other words, the real action is in, is the wave function for the center of mass of the hydrogen atom, and. So just what does this do for us? If in doubt, $$0^\circ \le \alpha ^\circ \textless 360^\circ$$, Dividing and factorising polynomial expressions, Solving logarithmic and exponential equations, Identifying and sketching related functions, Determining composite and inverse functions, Religious, moral and philosophical studies. For the sake of completeness we’ll close out this section with the 2-D and 3-D version of the wave equation. The initial conditions are then. Write $$2\sin x^\circ + 5\cos x^\circ$$ in the form $$k\sin (x + \alpha )^\circ$$ where $$k\textgreater0$$ and $$0^\circ \le \alpha ^\circ \textless 360^\circ$$. 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 $$k\sin \alpha ^\circ \textgreater 0$$. In the x,t (space,time) plane F(x − ct) is constant along the straight line x − ct = constant. In this section we want to consider a vertical string of length $$L$$ that has been tightly stretched between two points at $$x = 0$$ and $$x = L$$. This just means, make them equal each other. 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: What you will find is that you can actually ignore. Solving the Wave Function of R Using the Schrödinger Equation, Find the Eigenfunctions of Lz in Spherical Coordinates, Find the Eigenvalues of the Raising and Lowering Angular Momentum…, How Spin Operators Resemble Angular Momentum Operators. The addition formulae and trigonometric identities are used to simplify or evaluate trigonometric expressions. If we now divide by the mass density and define. Requiring the wave function to terminate at the right end of the tube gives , 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 force is called the tension in the string and its magnitude will be given by $$T\left( {x,t} \right)$$. 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, $k = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}$.