A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 m. The maximum transverse acceleration of a point at the middle of the segment is \(8.40 \times {10^3}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\) and the maximum transverse velocity is \(3.80\;{\rm{m/s}}\).

(a) What is the amplitude of this standing wave?

(b) What is the wave speed for the transverse travelling waves on this string?

Length of string \(L = 0.386\;m\).

Maximum transverse velocity \({v_{\max }} = 3.80\;{\rm{m/s}}\).

Maximum transverse acceleration \({a_{\max }} = 8.40 \times {10^3}\;{\rm{m/}}{{\rm{s}}^2}\).

The formula of maximum transverse velocity and maximum transverse acceleration.

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

Use the formula of maximum transverse velocity is calculated as follows:

\(\begin{array}{c}{v_{\max }} = A\omega \\A\omega = 3.80\;{\rm{m/s}}\;\;\;\;\;\;\;\;\;\;\;.....{\rm{(1)}}\end{array}\)

Use the formula of maximum transverse acceleration is calculated as follows:

\(\begin{array}{c}{a_{\max }} = A{\omega ^2}\\A{\omega ^2} = 8.4 \times {10^3}\;{\rm{m/}}{{\rm{s}}^2}\;\;\;\;\;\;\;\;\;\;\;.....{\rm{(2)}}\end{array}\)

Divide equation (2) by equation (1) gives:

\(\begin{array}{c}\frac{{A{\omega ^2}}}{{A\omega }} = \frac{{8.4 \times {{10}^3}{\rm{m/}}{{\rm{s}}^2}\;}}{{3.80\;{\rm{m/s}}}}\\\omega = \frac{{8.4 \times {{10}^3}{\rm{m/}}{{\rm{s}}^2}\;}}{{3.80\;{\rm{m/s}}}}\;\;\;\;\;.....(3)\end{array}\)

Substitute equation (3) in equation (1) gives:

\(\begin{array}{l}A\omega = 3.80\;{\rm{m/s}}\\A \times \frac{{8.4 \times {{10}^3}{\rm{m/}}{{\rm{s}}^2}\;}}{{3.80\;{\rm{m/s}}}} = 3.80\;{\rm{m/s}}\\A = \frac{{3.80\;{\rm{m/s}} \times 3.80\;{\rm{m/s}}}}{{8.4 \times {{10}^3}{\rm{m/}}{{\rm{s}}^2}}}\\A = 1.72 \times {10^{ - 3}}{\rm{m}}\end{array}\)

Hence, the amplitude of the standing wave is \(1.72 \times {10^{ - 3}}{\rm{m}}\).