A sinusoidal transverse wave travel on a string. The string has length \(8.00\;{\rm{m}}\) and mass \(6.00\;{\rm{g}}\). The wave speed is \(30.0\;{\rm{m/s}}\) and the wavelength is \(0.200\;{\rm{m}}\).

(a) If the wave is have an average power of \(50.0\;{\rm{W}}\), what must be the amplitude of the wave?

(b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

Mass of string \(m = 6.00\;{\rm{g}}\).

Length of string \(L = 8.00\;m\)

Speed of wave \(v = 30.0\;{\rm{m/s}}\).

Wavelength is \(\lambda = 0.2\;{\rm{m}}\)

The formula of power is:

\(\begin{array}{l}P = \rho {\omega ^2}{A^2}Sv\\A = \frac{1}{\omega }\sqrt {\frac{{2P}}{{\rho Sv}}} \end{array}\).

Here,\(\rho S = \mu \)is mass per unit length.

The angular velocity is calculated as follows:

\(\begin{array}{c}\omega = 2\pi f\\ = \frac{{2\pi v}}{\lambda }\\ = \frac{{2\pi }}{{0.2}} \times 30\end{array}\)

Also, calculate the mass per unit length as follows:

\(\begin{array}{c}\mu = \rho S\\ = \frac{{6 \times {{10}^{ - 3}}}}{8}{\rm{kg/m}}\end{array}\)

Substitute these values in the formula of amplitude of the wave gives:

\(\begin{array}{c}A = \frac{{0.2}}{{2\pi \times 30}}\sqrt {\frac{{2 \times 50 \times 8}}{{6 \times {{10}^{ - 3}} \times 30}}} \\ = 0.0707\;m\\ = 7.07\;cm\end{array}\)

Thus, the amplitude of the wave is 7.07 cm.