Question

A steel wire has a mass of \[5g/m\]and is under tension \[450N\]. Find the maximum average power that can be carried by the transverse wave in the wire if the amplitude is not to exceed $20\% $ of the wavelength.

Answer 1

Hint: Solution of this problem can found out by average power formula,
So, \[{P_{avg}} = 2{\pi ^2}\mu v{A^2}{f^2}\]
The time-averaged power of a sinusoidal wave on a string is found by average power= \[12\mu A^2\omega^2v,\] Average power = \[12\mu A^2\omega^2v,\] , where μ is the linear mass density of the string, A is the amplitude of the wave, ω is the angular frequency of the wave, and v is the speed of the wave.


Step by step answer: The power of the wave relies upon each amplitude and the frequency. If the power of every wavelength is taken into consideration to be a discrete packet of power, a high-frequency wave will supply greater of those packets per unit time than a low-frequency wave. We will see that the common rate of power transferred in mechanical waves is proportional to each square of the amplitude and the square of the frequency as well.
By solving with average power formula, we get
\[{P_{avg}} = 2{\pi ^2}\mu v{A^2}{f^2}\]
\[\Rightarrow {P_{avg}}\max = 2{\pi ^2}\mu v{\left( {0.2\lambda } \right)^2}{f^2}\]
\[\Rightarrow {P_{avg}}\max = 2{\pi ^2}\mu v{\left( {0.2} \right)^2}{\left( {f\lambda } \right)^2}\]
\[\Rightarrow {P_{avg}}\max = 2{\pi ^2}\mu {\left( {0.2} \right)^2}{v^3}\]
So, for calculating velocity we solve, \[v = \sqrt {\dfrac{T}{\mu }} = \sqrt {\dfrac{{450}}{{5 \times {{10}^{ - 3}}}} = } 300m/s\]
\[\therefore {P_{avg}}\max = 1080KW\]
Therefore, the Maximum Average Power is calculated as \[{P_{avg}}\max = 1080KW\]

Note: If mechanical waves have same amplitudes, however one wave has a frequency same to two times the frequency of the other, the higher-frequency wave may have a rate of power transfer a issue of 4 times as the exceptional because the rate of power transfer of the lower-frequency wave according to time period.