Question

A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density $6\,g/m$ kept under a tension of 60 N. Find the total energy associated with the wave in a 2.0 m long portion of the string.

Answer 1

Use the formula for the total energy stored in the string due to tension in the string. Convert the frequency of the wave in angular frequency wherever necessary.
Use the relation between wavelength, velocity and frequency to determine the wavelength of the wave.

Formula used:
$U = \dfrac{1}{2}\mu {\omega ^2}{A^2}\lambda$
Here, $\mu$ is the linear mass density of the string, $\omega$ is the angular frequency, A is the amplitude of the wave and $\lambda$ is the wavelength of the wave.

Complete step by step answer:
The energy associated with the wave is related to the amplitude and the frequency of the wave given by the expression,
$U = \dfrac{1}{2}\mu {\omega ^2}{A^2}\lambda$ …… (1)
Here, $\mu$ is the linear mass density of the string, $\omega$ is the angular frequency, A is the amplitude of the wave and $\lambda$ is the wavelength of the wave.
The angular frequency of the wave is given by the relation,
$\omega = 2\pi f$
Here, f is the frequency of the wave.
Calculate the angular frequency of the given wave as follows,
$\omega = 2\pi f$
Substitute $200\,Hz$ for f in the above equation.
$\omega = 2\pi \left( {200\,Hz} \right)$
$\therefore \omega = 1256\,rad/s$
Calculate the velocity of the wave over a string of linear mass density $\mu$ kept under the tension T as follows,
$v = \sqrt {\dfrac{T}{\mu }}$
Substitute 60 N for T and $6\,g/m$ for $\mu$ in the above equation.
$v = \sqrt {\dfrac{{60N}}{{\left( {6\,g/m} \right)\left( {\dfrac{{{{10}^{ - 3}}\,kg}}{{1\,g}}} \right)}}}$
$\Rightarrow v = \sqrt {\dfrac{{60\,N}}{{6 \times {{10}^{ - 3}}\,kg/m}}}$
$\Rightarrow v = \sqrt {10000\,{m^2}{s^{ - 2}}}$
$\therefore v = 100\,m/s$
We know the relation between velocity, wavelength and frequency of the wave. The wavelength of the wave is velocity divided by frequency of the wave.
Therefore,
$\lambda = \dfrac{v}{f}$
Substitute $100\,m/s$ for v and 200 Hz for f in the above equation.
$\lambda = \dfrac{{100\,m/s}}{{200\,Hz}}$
$\therefore \lambda = 0.5\,m$
Now, substitute 0.5 m for $\lambda$, $1256\,rad/s$ for $\omega$, 1 mm for A, and $6\,g/m$ for $\mu$ in equation (1) to calculate the energy of the wave.
$U = \dfrac{1}{2}\left( {6 \times {{10}^{ - 3}}\,kg/m} \right){\left( {1256\,rad/s} \right)^2}{\left( {\left( {1\,mm} \right)\left( {\dfrac{{{{10}^{ - 3}}\,m}}{{1\,mm}}} \right)} \right)^2}\left( {0.5\,m} \right)$
$\Rightarrow U = \dfrac{{4.74 \times {{10}^{ - 3}}}}{2}$
$\therefore U = 2.36 \times {10^{ - 3}}\,J$
This is the energy of one wavelength of the wave. We have to determine how many waves were actually there.
The total number of waves in the strings is equal to the length of the string divided by the wavelength of the wave.
$n = \dfrac{l}{\lambda }$
Substitute 2 m for l and 0.5 m for $\lambda$ in the above equation.
$n = \dfrac{{2\,m}}{{0.5\,m}}$
$\Rightarrow n = 4$
Therefore, the total energy of the wave is equal to the number of waves multiplied by the energy of the single wave.
$U = 4 \times 2.37 \times {10^{ - 3}}\,J$
$\Rightarrow U = 9.48 \times {10^{ - 3}}\,J$
Therefore, the energy associated with the wave is $9.48 \times {10^{ - 3}}\,J$.

Note: Always calculate the total number of waves produced in the string due to tension in the string. The energy associated with the wave is equal to energy of the single wavelength multiplied by the total number of wavelengths.