Question

A long string under tension of 100 N has one end at x = 0. A sinusoidal wave is generated at x = 0 whose equation is given by
$y=\left( 0.01cm \right)\sin \left[ \left( \dfrac{\pi x}{10}m \right)-50\pi t\sec \right]$
Find the average power transmitted by the wave.

Answer 1

Hint: We have a sinusoidal equation of wave as: $y=\left( 0.01cm \right)\sin \left[ \left( \dfrac{\pi x}{10}m \right)-50\pi t\sec \right]$
Compare the given equation with the standard form, i.e.: $y=A\sin \left[ kx-\omega t \right]$ and get the value of A, k and $\omega $.
Now, we need to find the average power transmitted by the wave. It is given by:
$\left\langle P \right\rangle =\dfrac{1}{2}{{\omega }^{2}}{{A}^{2}}\sqrt{T\mu }$, where \[\mu =\dfrac{T}{{{v}^{2}}}\] , $v=\dfrac{\omega }{k}$ and \[T=\dfrac{2\pi }{\omega }\]. Put all the values in the formula and find the average power transmitted.

Complete step by step answer:
We have a sinusoidal equation of wave as: $y=\left( 0.01cm \right)\sin \left[ \left( \dfrac{\pi x}{10}m \right)-50\pi t\sec \right]......(1)$
By comparing equation (1) with $y=A\sin \left[ kx-\omega t \right]$, we get:
$\begin{align}
  & A=0.01cm={{10}^{-4}}m \\
 & k=\dfrac{\pi }{10}m \\
 & \omega =50\pi \sec \\
\end{align}$
Now, using the above data, find the value of $\mu ,v\text{ and }T$
We get:
\[\begin{align}
  & T=\dfrac{2\pi }{\omega } \\
 & =\dfrac{2\pi }{50\pi } \\
 & =\dfrac{1}{25}\sec
\end{align}\]
$\begin{align}
  & v=\dfrac{50\pi }{\dfrac{\pi }{10}} \\
 & =500m/s
\end{align}$
\[\begin{align}
  & \mu =\dfrac{\dfrac{1}{25}}{{{\left( 500 \right)}^{2}}} \\
 & =\dfrac{1}{125\times {{10}^{4}}}{{m}^{-2}}{{s}^{2}}
\end{align}\]
Now, put all the values to find average power transmitted by the wave.
We get:
$\begin{align}
  & \left\langle P \right\rangle =\dfrac{1}{2}{{\left( 50\pi \right)}^{2}}{{\left( {{10}^{-4}} \right)}^{2}}\sqrt{\left( \dfrac{1}{25} \right)\left( \dfrac{1}{125\times {{10}^{4}}} \right)} \\
 & =4.93\times {{10}^{-8}}W
\end{align}$\[\begin{align}
  & \mu =\dfrac{\dfrac{1}{25}}{{{\left( 500 \right)}^{2}}} \\
 & =\dfrac{1}{125\times {{10}^{4}}}{{m}^{-2}}{{s}^{2}}
\end{align}\]

Note:
The energy associated with a traveling wave in a stretched string is conveniently expressed as the energy per wavelength. Since this amount of energy is transported a distance of one wavelength along the string in one period, this expression can be used to calculate the power transmitted along a string.
So, the power transmitted by a string is given by the formula: $\left\langle P \right\rangle =\dfrac{1}{2}{{\omega }^{2}}{{A}^{2}}\sqrt{T\mu }$.