A strong string of mass 3.00 g and length 2.20 m is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is \(9.00\;{\rm{m/s}}\). The tension in the string is 330 N.

(a) What is the amplitude of the standing wave at its antinode?

(b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

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 \({\rm{1}}{\rm{.281}}\;c{\rm{m}}\).