Answer
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Hint In this solution, we will use the general form of the wave equation and we will then simplify the wave equation to the form of the wave equation given in the question. The phase difference between two particles can be obtained from the relation of the phase difference with the path difference of the two particles in the wave.
Formula used: In this question, we will use the following formula of the wave function,
$\Rightarrow y = A\sin (\omega t - kx)$ where $A$ is the amplitude, $\omega $ is the angular frequency, $t$ is the time, $k$ is the wave number, and $x$ is the displacement of a given point in the wave from its mean position.
Complete step by step answer
The general form of the wave function is
$\Rightarrow y = A\sin (\omega t - kx)$
Now we know that $\omega = 2\pi f$ and $k = 2\pi /\lambda $, we can substitute these values in the wave function as:
$\Rightarrow y = A\sin \left( {2\pi ft - \dfrac{{2\pi }}{\lambda }x} \right)$
Taking $2\pi $ common, we get
$\Rightarrow y = A\sin 2\pi \left( {ft - \dfrac{1}{\lambda }x} \right)$
This form is similar to the form of the wave function in the question $y = 8sin2\pi \left( {0.1x - 2t} \right)$. So comparing these two waves forms, we get,
$\Rightarrow \dfrac{1}{\lambda } = 0.1$
$\Rightarrow \lambda = 10\,cm$
Then, we can calculate the path difference between two points $\Delta x = 2\,cm$ apart as
$\Rightarrow \Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta x$
$\Rightarrow \Delta \phi = \dfrac{{2\pi }}{{10}}(2)$
Thus, the phase difference is
$\Delta \phi = 72^\circ $ which corresponds to option (D).
Note
The phase difference between two particles in the wave relates to the relative displacement of two particles from the mean position in terms of angels/radians. This phase difference between two points is constant with time if the wave is travelling at a constant wavelength however the position of the two particles change with respect to the mean position in such a way that the phase difference of the two particles remains constant.
Formula used: In this question, we will use the following formula of the wave function,
$\Rightarrow y = A\sin (\omega t - kx)$ where $A$ is the amplitude, $\omega $ is the angular frequency, $t$ is the time, $k$ is the wave number, and $x$ is the displacement of a given point in the wave from its mean position.
Complete step by step answer
The general form of the wave function is
$\Rightarrow y = A\sin (\omega t - kx)$
Now we know that $\omega = 2\pi f$ and $k = 2\pi /\lambda $, we can substitute these values in the wave function as:
$\Rightarrow y = A\sin \left( {2\pi ft - \dfrac{{2\pi }}{\lambda }x} \right)$
Taking $2\pi $ common, we get
$\Rightarrow y = A\sin 2\pi \left( {ft - \dfrac{1}{\lambda }x} \right)$
This form is similar to the form of the wave function in the question $y = 8sin2\pi \left( {0.1x - 2t} \right)$. So comparing these two waves forms, we get,
$\Rightarrow \dfrac{1}{\lambda } = 0.1$
$\Rightarrow \lambda = 10\,cm$
Then, we can calculate the path difference between two points $\Delta x = 2\,cm$ apart as
$\Rightarrow \Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta x$
$\Rightarrow \Delta \phi = \dfrac{{2\pi }}{{10}}(2)$
Thus, the phase difference is
$\Delta \phi = 72^\circ $ which corresponds to option (D).
Note
The phase difference between two particles in the wave relates to the relative displacement of two particles from the mean position in terms of angels/radians. This phase difference between two points is constant with time if the wave is travelling at a constant wavelength however the position of the two particles change with respect to the mean position in such a way that the phase difference of the two particles remains constant.
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