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Hint: To find out the relation between the rate transfer of energy in a wave and the wave amplitude use formula for power ${{ P = }}\dfrac{{{1}}}{{{2}}}{{\mu }}{{{A}}^{{2}}}{{{\omega }}^{{2}}}$. As rate transfer of energy is equal to the power i.e. $\dfrac{{{{dE}}}}{{{{dt}}}}{{ = P}}$. Similarly using the same formula, find out the relation between the rate transfer of energy in a wave and the wave frequency. So, both the relations can be found using the same formula.
Complete step by step solution:
Wave motion is a disturbance which travels through a medium because of the repeated vibrations of the particles of the medium about their mean positions, in actual the disturbance is shifted from one particle to the next particle and so on.
The rate transfer of energy $\dfrac{{{{dE}}}}{{{{dt}}}}$ is also termed as power P.
Formula for time-average power is given by
$\dfrac{{{{dE}}}}{{{{dt}}}}{{ = P = }}\dfrac{{{1}}}{{{2}}}{{\mu }}{{{A}}^{{2}}}{{{\omega }}^{{2}}}$
Where ${{\mu = }}$Mass per unit length of the wave
A = Amplitude of the wave
${{\omega = }}$Wave frequency
Here $\dfrac{{{{dE}}}}{{{{dt}}}}{{ \propto }}{{{A}}^{{2}}}$
Thus, the rate transfer of energy in a wave depends directly on the square of the wave amplitude.
Also, $\dfrac{{{{dE}}}}{{{{dt}}}}{{ \propto }}{{{\omega }}^{{2}}}$
So, the rate transfer of energy in a wave depends directly on the square of the wave frequency.
Therefore, option (A) is the correct choice.
Note: When a transverse or longitudinal wave propagates through a medium, all the particles of the medium oscillate about the mean positions in the same manner but phase of oscillation changes from one particle to the next particle. Amplitude is the maximum displacement suffered by the particles of the medium about their mean positions. It is denoted by A. Angular frequency is the rate of phase with time. It is denoted by ${{\omega }}$.
Complete step by step solution:
Wave motion is a disturbance which travels through a medium because of the repeated vibrations of the particles of the medium about their mean positions, in actual the disturbance is shifted from one particle to the next particle and so on.
The rate transfer of energy $\dfrac{{{{dE}}}}{{{{dt}}}}$ is also termed as power P.
Formula for time-average power is given by
$\dfrac{{{{dE}}}}{{{{dt}}}}{{ = P = }}\dfrac{{{1}}}{{{2}}}{{\mu }}{{{A}}^{{2}}}{{{\omega }}^{{2}}}$
Where ${{\mu = }}$Mass per unit length of the wave
A = Amplitude of the wave
${{\omega = }}$Wave frequency
Here $\dfrac{{{{dE}}}}{{{{dt}}}}{{ \propto }}{{{A}}^{{2}}}$
Thus, the rate transfer of energy in a wave depends directly on the square of the wave amplitude.
Also, $\dfrac{{{{dE}}}}{{{{dt}}}}{{ \propto }}{{{\omega }}^{{2}}}$
So, the rate transfer of energy in a wave depends directly on the square of the wave frequency.
Therefore, option (A) is the correct choice.
Note: When a transverse or longitudinal wave propagates through a medium, all the particles of the medium oscillate about the mean positions in the same manner but phase of oscillation changes from one particle to the next particle. Amplitude is the maximum displacement suffered by the particles of the medium about their mean positions. It is denoted by A. Angular frequency is the rate of phase with time. It is denoted by ${{\omega }}$.
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