A simple harmonic progressive wave of friction $5Hz$ is traveling along the positive direction with a velocity of$40m{{s}^{-1}}$. Calculate the phase difference between two points separated by a $0.8m$ distance.
Answer
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Hint: Phase Difference is used to explain the difference in degrees or radians when two or more alternating quantities reach their maximum or minimum values. In this question equation containing wavelength, frequency and velocity is being used. By that way we can find the path difference and then its phase difference. If the path difference is $\lambda $, then the phase difference will be given as $2\pi $.
Complete step by step solution: first of all let us discuss the path difference and phase difference. Path difference abbreviated as PD is the difference in the distance traversed by the two waves from their own sources in order to give a point a pattern. The path difference is equal to one wavelength. Phase Difference is used to explain the difference in degrees or radians when two or more alternating quantities reach their maximum or minimum values. The phase difference can be measured in degree or radian while path difference is measured in angstrom. If the path difference is$\lambda $, then the phase difference will be given as $2\pi $ .
In this question it is given that,
$\begin{align}
& f=5Hz \\
& v=40m{{s}^{-1}} \\
\end{align}$
As we all know,
$\lambda =\dfrac{v}{f}$
Substituting the values in it,
$\lambda =\dfrac{40}{5}=8m$
As we know the path difference is given by the formula,
$\Delta \phi =\dfrac{2\pi }{\Delta \lambda }$
Substituting the value of path difference given in the question here in the $\Delta \phi =\dfrac{\pi }{5}rad$ is will give,
$0.8=\dfrac{2\pi }{\Delta \phi }$
Therefore
$\Delta \phi =\dfrac{\pi }{5}rad$
So the phase difference will be
$\Delta \phi =\dfrac{\pi }{5}rad$
Hence we got the answer.
Note: simple harmonic motion is a special case in periodic motion in which the restoring force of the object in motion is directly proportional to the displacement of the object, magnitude and its act towards the equilibrium position of the object.
Complete step by step solution: first of all let us discuss the path difference and phase difference. Path difference abbreviated as PD is the difference in the distance traversed by the two waves from their own sources in order to give a point a pattern. The path difference is equal to one wavelength. Phase Difference is used to explain the difference in degrees or radians when two or more alternating quantities reach their maximum or minimum values. The phase difference can be measured in degree or radian while path difference is measured in angstrom. If the path difference is$\lambda $, then the phase difference will be given as $2\pi $ .
In this question it is given that,
$\begin{align}
& f=5Hz \\
& v=40m{{s}^{-1}} \\
\end{align}$
As we all know,
$\lambda =\dfrac{v}{f}$
Substituting the values in it,
$\lambda =\dfrac{40}{5}=8m$
As we know the path difference is given by the formula,
$\Delta \phi =\dfrac{2\pi }{\Delta \lambda }$
Substituting the value of path difference given in the question here in the $\Delta \phi =\dfrac{\pi }{5}rad$ is will give,
$0.8=\dfrac{2\pi }{\Delta \phi }$
Therefore
$\Delta \phi =\dfrac{\pi }{5}rad$
So the phase difference will be
$\Delta \phi =\dfrac{\pi }{5}rad$
Hence we got the answer.
Note: simple harmonic motion is a special case in periodic motion in which the restoring force of the object in motion is directly proportional to the displacement of the object, magnitude and its act towards the equilibrium position of the object.
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