
\[100\pi \] phase difference$=$_____ path difference
$\begin{align}
& \text{A}\text{. 10}\lambda \\
& \text{B}\text{. 25}\lambda \\
& \text{C}\text{. 50}\lambda \\
& \text{D}\text{. 100}\lambda \\
\end{align}$
Answer
518.4k+ views
Hint: A wave is represented by sinusoidal form. So at the time of measurement I.e. $t=0$ it does not pass through the origin and said to have a phase difference or phase shift. The phase difference or phase shift is the angle $\phi $ in degrees or radians that the waveform has shifted from a certain reference point along the horizontal zero axis. Path difference is the difference between the path travelled of two waves at a certain point.
Formula used:
Relationship between the phase difference$\left( \Delta \phi \right)$ , path difference$\left( \Delta x \right)$ and the wavelength $\left( \lambda \right)$of a wave is given by $\Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x$
Complete answer:
Here the phase difference given is $100\pi $ i.e. $\Delta \phi =100\pi $
But the relationship between the phase difference$\left( \Delta \phi \right)$ , path difference$\left( \Delta x \right)$ and the wavelength $\left( \lambda \right)$of a wave is given by
$\begin{align}
& \Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x,\text{ so } \\
& \Rightarrow 100\pi =\dfrac{2\pi }{\lambda }\times \Delta x \\
& \Rightarrow \Delta x=50\lambda \\
\end{align}$
So, the correct answer is “Option C”.
Additional Information:
The equation of wave motion is given by
$y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)$
where
$\begin{align}
& y\left( x,t \right)=\text{ displacement} \\
& A=\text{ amplitude} \\
& \omega =\text{ angular frequency} \\
& t=\text{time} \\
& k=\text{angular wave number} \\
& x=\text{position} \\
& {{\phi }_{0}}=\text{ initial phase} \\
\end{align}$
Phase: The phase of a harmonic wave is a quantity that gives complete information of the wave at any time and any position.
If a wave is represented by $y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)$
Then phase of the wave at position $x$ and time$t$ is given by. $\phi =\left( \omega t-kx+{{\phi }_{0}} \right)$
So clearly the phase of a wave is periodic both in time and space. So at a given point $x$the phase changes with time and at a given time $t$ the phase changes with position $x$
Now, phase ,$\phi =\left( \omega t-kx+{{\phi }_{0}} \right)$
Taking $x$as a constant differentiate $\phi$ with respect to $t$ then,
$\dfrac{\Delta \phi }{\Delta t}=\omega $
Thus the phase change at a given position $x$ in time $\Delta t$ is given by
$\Delta \phi =\omega \Delta t=\dfrac{2\pi }{T}\Delta t$
Where T= Time period of the wave.
Taking time as constant
$\begin{align}
& \dfrac{\Delta \phi }{\Delta x}=-k \\
& \Rightarrow \Delta \phi =-k\Delta x=-\dfrac{2\pi }{\lambda }\Delta x \\
\end{align}$
Note:
Note that from the equation of wave motion you can calculate the particle velocity particle acceleration, particle energy. Generally waves are of three main types such as.
Mechanical wave-it is the wave which obeys Newton’s law and can only exist within material medium e.g. water wave, sound wave, seismic wave.
Electromagnetic wave- wave associated with a charged objects.e.g x-ray.
Matter waves-waves associated with fundamental particles like electron,proton and neutron.
Formula used:
Relationship between the phase difference$\left( \Delta \phi \right)$ , path difference$\left( \Delta x \right)$ and the wavelength $\left( \lambda \right)$of a wave is given by $\Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x$
Complete answer:
Here the phase difference given is $100\pi $ i.e. $\Delta \phi =100\pi $
But the relationship between the phase difference$\left( \Delta \phi \right)$ , path difference$\left( \Delta x \right)$ and the wavelength $\left( \lambda \right)$of a wave is given by
$\begin{align}
& \Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x,\text{ so } \\
& \Rightarrow 100\pi =\dfrac{2\pi }{\lambda }\times \Delta x \\
& \Rightarrow \Delta x=50\lambda \\
\end{align}$
So, the correct answer is “Option C”.
Additional Information:
The equation of wave motion is given by
$y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)$
where
$\begin{align}
& y\left( x,t \right)=\text{ displacement} \\
& A=\text{ amplitude} \\
& \omega =\text{ angular frequency} \\
& t=\text{time} \\
& k=\text{angular wave number} \\
& x=\text{position} \\
& {{\phi }_{0}}=\text{ initial phase} \\
\end{align}$
Phase: The phase of a harmonic wave is a quantity that gives complete information of the wave at any time and any position.
If a wave is represented by $y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)$
Then phase of the wave at position $x$ and time$t$ is given by. $\phi =\left( \omega t-kx+{{\phi }_{0}} \right)$
So clearly the phase of a wave is periodic both in time and space. So at a given point $x$the phase changes with time and at a given time $t$ the phase changes with position $x$
Now, phase ,$\phi =\left( \omega t-kx+{{\phi }_{0}} \right)$
Taking $x$as a constant differentiate $\phi$ with respect to $t$ then,
$\dfrac{\Delta \phi }{\Delta t}=\omega $
Thus the phase change at a given position $x$ in time $\Delta t$ is given by
$\Delta \phi =\omega \Delta t=\dfrac{2\pi }{T}\Delta t$
Where T= Time period of the wave.
Taking time as constant
$\begin{align}
& \dfrac{\Delta \phi }{\Delta x}=-k \\
& \Rightarrow \Delta \phi =-k\Delta x=-\dfrac{2\pi }{\lambda }\Delta x \\
\end{align}$
Note:
Note that from the equation of wave motion you can calculate the particle velocity particle acceleration, particle energy. Generally waves are of three main types such as.
Mechanical wave-it is the wave which obeys Newton’s law and can only exist within material medium e.g. water wave, sound wave, seismic wave.
Electromagnetic wave- wave associated with a charged objects.e.g x-ray.
Matter waves-waves associated with fundamental particles like electron,proton and neutron.
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