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**[SOLVED] Standing Waves on a String with a Free End**

Problem.A string can have a "free" end if that end is attached to a ring that can slide without friction on a vertical pole. Determine the wavelengths of the resonant vibrations of such a string with one end fixed and the other free.

Problem.

By resonant vibrations, I guess they mean standing waves. A standing wave is the superposition of two waves traveling in opposite direction right? The general equation of a traveling wave is given by

[tex]D(x,t) = D_M \sin (kx \pm \omega t + \phi)[/tex]

Suppose the pole is on the left side. Let L be the distance from the pole to the fixed end. To create waves, one would have to move the ring up and down at some particular frequency right? That means that the equation of the wave traveling to the left has a non-zero phase angle [itex]\phi[/itex]. Let

[tex]D_1 = D_M \sin (kx - \omega t + \phi)[/tex]

be the equation of this wave. The reflected wave is the same as the incident wave except that it is traveling to the right and since it starts from a fixed location, its phase angle is zero. Let

[tex]D_2 = D_M \sin (kx + \omega t)[/tex]

be the equation of this wave. The equation of the standing wave is

[tex]D = D_1 + D_2 = 2D_M \sin (kx + \phi/2) \cos (\phi/2 - \omega t)[/tex]

D = 0 at x = L which means that [itex]kL + \phi/2 = n \pi[/itex] for some non-negative integer n or

[tex]\lambda = \frac{4 \pi L}{2\pi n - \phi}[/tex]

My book has as its answer 4L/(2n - 1). It's as if they decided to make [itex]\phi = \pi[/itex]. Why is that?

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