91Ó°ÊÓ

1. Length of string,$I=2.7\mathrm{m}$
2. Mass of string, $m=260\mathrm{g}=0.260\mathrm{kg}$
3. Tension in the string, $T=36\mathrm{N}$
4. Amplitude of waves, $y_m=7.70\mathrm{mm}=7.70×{10}^{-3}\mathrm{m}$
5. The average power,$P_{ave}=85.0\mathrm{W}$

When we set up a wave on a stretched string, as the wave moves away from us, it transports both kinetic as well as potential energy. The rate of transfer of energy along the string gives power.

The average power of a body,

$P_{ave}=\frac{1}{2}\mathrm{μ}\mathrm{ν}{\mathrm{Ó\neg }}^2{y}_m^2$ (i)

The angular frequency of a body,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ (ii)

The velocity of a body,

$v=\sqrt{\frac{T}{\mathrm{μ}}}$ (iii)

The linear density of a body distribution,

$\mathrm{μ}=\frac{m}{I}$ (iv)

Here, $y_m$is the amplitude, $T$is the tension in the string, $I$is the length of the string and $f$ is the frequency of oscillation.

From equation (iv), the linear density is given as:

$$\begin{gathered} \mathrm{μ}=\frac{0.260\mathrm{g}}{2.7\mathrm{m}} \\ =0.096\mathrm{g}/\mathrm{m} \end{gathered}$$

Next, we have to use the following formula to calculate wave speed

$\mathrm{ν}=\sqrt{\frac{\mathrm{τ}}{\mathrm{μ}}}$

Rearranging the equation (i) of the average power formula, we can get

$$\begin{gathered} {\mathrm{Ó\neg }}^2=\frac{2P_{avg}}{\mathrm{μ}\mathrm{ν}{y}_m^2} \\ {f}^2=\frac{2P_{avg}}{4{\mathrm{Ï€}}^2\mathrm{μ}\mathrm{ν}{y}_m^2}\left(âˆ\mu \mathrm{using}\mathrm{equation}\left(\mathrm{ii}\right)\right) \\ f=\sqrt{\frac{2P_{avg}}{4{\mathrm{Ï€}}^2\mathrm{μ}\mathrm{ν}{y}_m^2}} \\ f=\frac{1}{{\mathrm{Ï€²â}}_{\mathrm{m}}}\sqrt{\frac{2P_{avg}}{2\mathrm{μ}\sqrt{{\displaystyle \frac{T}{\mathrm{μ}}}}}}\left(âˆ\mu \mathrm{using}\mathrm{equation}\left(\mathrm{iii}\right)\right) \end{gathered}$$

On solving further,

$$\begin{gathered} f=\frac{1}{\mathrm{π}{y}_m}\sqrt{\frac{P_{avg}}{2\sqrt{T\mathrm{μ}}}} \\ =\frac{1}{3.14×\left(7.70×{10}^{-3}\right)\mathrm{m}}\sqrt{\frac{85.0\mathrm{W}}{2\sqrt{36.0\mathrm{N}×0.096\mathrm{kg}/\mathrm{m}}}} \\ =41.36{\mathrm{m}}^{-1}×\sqrt{\frac{85.0\mathrm{W}}{2\sqrt{36.0\mathrm{N}×0.096\mathrm{kg}/\mathrm{m}}}} \\ =198\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency of traveling waves of amplitude 7.70 mm for the average power to be 85.0 w , is 198 Hz .