Question

A string can have a “free” end if that end is attached to a ring that can slide without friction on a vertical pole (Fig. 11–60). Determine the wavelengths of the resonant vibrations of such a string with one end fixed and the other free.

Short answer

The wavelength of the resonant vibrations is given by \(\lambda \; = \;4l\left( {\frac{1}{{2n - 1}}} \right)\).

Step-by-step solution

Given data in the question

One end of the string is fixed, and the other end is free. The free end can slide without friction on the vertical pole.

The length of the string is \(l\).

Concepts used in the solution

In a stationary wave, there are nodes and antinodes. Nodes are the points of zero displacements, and antinodes are the points with maximum displacements.

The distance between a node and an antinode is\(\frac{\lambda }{4}\).

The string can be assumed to be a closed pipe. For resonance vibrations, the fixed end of the string should be a node, and the free end should be an antinode.

Calculation for determining the desired result

The distance from fixed end to free end for the first resonance vibration is when \(l\; = \;\frac{\lambda }{4}\). Therefore, in this case, \(\lambda = 4l\).

The second resonance vibration occurs when \(l = \;\frac{\lambda }{2} + \;\frac{\lambda }{4} = \;\frac{{3\lambda }}{4}\). Therefore, \(\lambda = \;\frac{{4l}}{3}\).

The third resonance vibration occurs when \(l = \;\lambda + \;\frac{\lambda }{4} = \;\frac{{5\lambda }}{4}\). Therefore, \(\lambda = \;\frac{{4l}}{5}\).

The resonance pattern continues as above.

From the above calculations, it is clear that the wavelengths for resonance are,

\(\lambda = \;4l,\;\frac{{4l}}{3},\;\frac{{4l}}{5},\;\frac{{4l}}{7},......\; = \;4l\left( {1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ....} \right)\)

In general, we can write

\(\lambda \; = \;4l\left( {\frac{1}{{2n - 1}}} \right),\;n\; = \;1,\;2,\;3,\;....\)

The above expression gives the wavelengths of resonant vibrations.