91Ó°ÊÓ

Length of string \(L = 2.20\;m\).

The tension on the string \(T = 330.0\;{\rm{N}}\).

Mass of the string \(m = 3.00\;{\rm{g}}\).

Maximum transverse speed \({v_{\max }} = 9.00\;{\rm{m/s}}\).

The formula of maximum transverse velocity and maximum transverse acceleration are:

\({v_{\max }} = A\omega \)and\({a_{\max }} = A{\omega ^2}\)

The formula of mass is:

\(m = \mu L\)

Calculate the mass per unit length as follows:

\(\begin{array}{c}\mu = \frac{m}{L}\\ = \frac{{3.00\;g}}{{\left( {2.20\;{\rm{m}}} \right)}}\\ = 1.36 \times {10^{ - 3}}\;{\rm{kg/m}}\end{array}\)

The wavelength of the stationary wave is related to the length of the wire by the relation:

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

Here, \(n\) is number of nodes in the wire.

For fundamental node, \(n = 1\) thus,

\(\begin{array}{c}\lambda = 2L\\ = 2 \times 2.20\;{\rm{m}}\\ = 4.4\;{\rm{m}}\end{array}\)

The speed of the wave is,

\(\begin{array}{c}v = \sqrt {\frac{T}{\mu }} \\ = \sqrt {\frac{{330\;N \times 2.2\;m}}{{3 \times {{10}^{ - 3}}\,kg}}} \\ = \sqrt {242000} \\ = 491.9\\ = 492\;{\rm{m/s}}\end{array}\)

By the use of speed of the wave, the angular velocity is calculated as follows:

\(\begin{array}{c}\omega = 2\pi f\\ = 2\pi \frac{v}{\lambda }\\ = \frac{{2\pi \times 492\;{\rm{m/s}}}}{{4.4\;{\rm{m}}}}\\ = 702.5\;{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

Thus, the amplitude of the string is:

\(\begin{array}{c}A = \frac{{{v_{\max }}}}{\omega }\\ = \frac{9}{{702.5}}\\ = 0.01281\;{\rm{m}}\\{\rm{ = 1}}{\rm{.281}}\;c{\rm{m}}\end{array}\)

Hence, the amplitude of the string is \({\rm{1}}{\rm{.281}}\;c{\rm{m}}\).