91Ó°ÊÓ

The given data can be listed below as,

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

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, \(1.33\;{\rm{m}}\).