# Standing Waves on a String with a Free End

[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.

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

$$D(x,t) = D_M \sin (kx \pm \omega t + \phi)$$

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 $\phi$. Let

$$D_1 = D_M \sin (kx - \omega t + \phi)$$

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

$$D_2 = D_M \sin (kx + \omega t)$$

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

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

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

$$\lambda = \frac{4 \pi L}{2\pi n - \phi}$$

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

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Problem. 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

$$D_2 = D_M \sin (kx + \omega t)$$

be the equation of this wave.
When a wave is reflected it suffers a phase change of 'pi'. Hence the value of phi is 'pi'.

Not always, if the medium is free, the just the direction changes, if it isnt, then the phase change also takes place.

air to water is not a free medium boundary. water to air is. (denser->rarer).

oh! sorry. I forgot to tell : pi phase change occurs when wave reflects from rarer to denser interface

When a wave is reflected it suffers a phase change of 'pi'. Hence the value of phi is 'pi'.
What does the phase change of the reflected wave have to do with the incident wave's phase shift? How is this evident from the equations?

What does the phase change of the reflected wave have to do with the incident wave's phase shift? How is this evident from the equations?
I think I understand now. Given the equation of the reflected wave:

$$D_2 = D_M \sin (kx + \omega t)$$

the incident wave will have a phase shift of 180 degrees or $\pi$ and will travel in the opposite direction. Hence, its equation must be:

$$D_1 = D_M \sin (kx - \omega t + \pi)$$

Now, given the equation of the incident wave:

$$D_1 = D_M \sin (kx - \omega t + \phi)$$

the equation of the reflected wave must be

$$D_2 = D_M \sin (kx + \omega t + \phi + \pi)$$

and since the phase shift of this equation must be 0 (because of the fixed end), then $\phi + \pi = 0$. Hence $\phi = -\pi$. Hmm...but this is negative though.

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$$D_2 = D_M \sin (kx + \omega t + \phi + \pi)$$

and since the phase shift of this equation must be 0 (because of the fixed end), then $\phi + \pi = 0$. Hence $\phi = -\pi$. Hmm...but this is negative though.
It just realized, after fiddling with the equation a bit, that it doesn't matter if the phase shift is negavite. The wavelengths of the resonant vibrations will still be 4L/(2n - 1).