A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of \(28.0\;{\rm{m/s}}\).

(a) What is the string’s mass?

(b) What is the magnitude of the maximum transverse acceleration of this point on the string?

Length of string \(L = 2.50\;m\).

The tension on the string \(T = 90.0\;{\rm{N}}\).

The first overtone is \(L = \lambda \)

Amplitude of any point on the string \(A = 3.50\;c{\rm{m}}\).

Maximum transverse speed \({v_{\max }} = 2.80\;{\rm{m/s}}\).

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:

\(m = \mu L\)

The first overtone is \(L = \lambda \)

By the use of maximum transverse velocity, the angular velocity is calculated as follows:

\(\begin{array}{c}\omega = \frac{{{v_{\max }}}}{A}\\ = \frac{{2.80\;{\rm{m/s}}}}{{3.50\;{\rm{cm}}}}\\ = \frac{{2.80\;{\rm{m/s}}}}{{0.0350\;{\rm{m}}}}\\ = 700\;{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

The speed of the wave is,

\(\begin{array}{c}v = \sqrt {\frac{T}{\mu }} \\ = \frac{\omega }{k}\\ = \frac{{\lambda \omega }}{{2\pi }}\end{array}\)

Thus, the mass per unit length is written as:

\(\begin{array}{c}\mu = \frac{{4{\pi ^2}T}}{{{\lambda ^2}{\omega ^2}}}\\ = \frac{{4{\pi ^2} \times 90.0\;N}}{{{{\left( {2.50\;{\rm{m}}} \right)}^2}{{\left( {700\;{{\rm{s}}^{{\rm{ - 1}}}}} \right)}^2}}}\\ = 1.16 \times {10^{ - 3}}\;{\rm{kg/m}}\end{array}\)

Now, the mass of the string is:

\(\begin{array}{c}m = \mu L\\m = 1.16 \times {10^{ - 3}}\;{\rm{kg/m}} \times {\rm{2}}{\rm{.50}}\;{\rm{m}}\\ = 2.9\;{\rm{g}}\end{array}\)

Hence, the mass of the string is \(2.9\;{\rm{g}}\).