Waves on a Stick. A flexible stick \(2.0\;{\rm{m}}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

Short Answer

Expert verified

The wavelength in 1st, 2nd and 3rd harmonic are, \(4.0\;{\rm{m}}\), \(2.0\;{\rm{m}}\) and \(1.33\;{\rm{m}}\).

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The length of stick is, \(L = 2.0\;{\rm{m}}\).
02

Significance of the wavelength

The difference between similar points (adjacent crests) in successive cycles of a waveform signal travelling in space or over a wire.

03

Determination of the wavelengths

The stick's free ends can only be displacement antinodes because they are free. Every harmonic after the first has one node located in the middle of the stick. The distance between nodes and between antinodes is \(\frac{\lambda }{2}\) here \(\lambda \) is the wavelength.

The figure below shows the standing wave patterns for the first three harmonics.

For first harmonic;

The length is expressed as,

\(L = \frac{{{\lambda _1}}}{2}\)

Rearranging above equation,

\({\lambda _1} = 2L\) …(1)

Substitute the value of \(L\) in equation (1)

\(\begin{array}{l}{\lambda _1} = 2 \times 2.0\;{\rm{m}}\\{\lambda _1} = 4.0\;{\rm{m}}\end{array}\)

Hence the wavelength in first harmonic is, \(4.0\;{\rm{m}}\).

For second harmonic;

The length is expressed as,

\(L = {\lambda _2}\) …(2)

Substitute the value of \(L\) in equation (2)

\({\lambda _2} = 2.0\;{\rm{m}}\)

Hence the wavelength in second harmonic is, \(2.0\;{\rm{m}}\).

For third harmonic;

The length is expressed as,

\(L = \frac{3}{2}{\lambda _3}\)

Rearranging above equation,

\({\lambda _3} = \frac{2}{3}L\) …(3)

Substitute the value of \(L\) in equation (3)

\(\begin{array}{l}{\lambda _3} = \frac{2}{3} \times 2.0\;{\rm{m}}\\{\lambda _3} = 1.33\;{\rm{m}}\end{array}\)

Hence the wavelength in third harmonic is, \(1.33\;{\rm{m}}\).

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