The phase difference between two points is $\dfrac{\pi}{3}$. If the frequency of the wave is 50Hz, then what is the distance between two points? (Given v = 330m/s)
A). 2.2m
B). 1.1m
C). 0.6m
D). 1.7m
Answer
604.8k+ views
Hint: The phase difference between the two sound waves of the same frequency moving past a fixed location is given by the time difference between the same positions within the wave cycles of the two sounds.
The path difference is that difference that is travelled by the two waves from their respective sources to a give point on the pattern
Complete step by step solution:
The expression for the path difference and the phase difference
$\begin{align}
&Phase\;difference = \dfrac{{2\pi }}{\lambda } \times path\;difference \\
& \Delta \phi = \dfrac{{2\pi }}{\lambda } \times \Delta x \\
\end{align} $
Here, $\Delta \phi $ is the phase difference,
$\Delta x$ Is the path difference
$\lambda $ Is the wavelength
The phase difference between any two particles in a wave determines lack of harmony in the vibrating state of the two particles. i.e., how far one particle leads the other or lags behind the other.
From the relation,
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \Delta x$
$\Delta x = \dfrac{\lambda }{{2\pi }} \times \Delta \phi $ --- (1)
Also, $\lambda = \dfrac{v}{f}$ --- (2)
Here v is the velocity
f is the frequency
We know the value of the frequency and velocity, and now, we calculate the wavelength, from equation (2)
$\begin{align}
&\lambda = \dfrac{{\left( {330m/s} \right)}}{{\left( {50Hz} \right)}} \\
&\Rightarrow \lambda = 6.6m \\
\end{align} $
The wavelength is 6.6m
Now, we have the value of wavelength and the value of phase difference.
We calculate the path difference from an equation (1)
$\begin{align}
&\Delta x = \dfrac{\lambda }{{2\pi }} \times \Delta \phi \\
& \Rightarrow \Delta x = \left( {\dfrac{{6.6}}{{2 \times \pi }}} \right) \times \left( {\dfrac{\pi }{3}} \right) \\
& \Rightarrow \Delta x = \dfrac{{2.2}}{2} \\
& \Rightarrow \Delta x = 1.1m \\
\end{align} $
So, the value of the path difference is 1.1m
So, the option (B) is correct.
Note: The term phase difference, it is the difference in the phase angle of the two waves.
The term path difference, it is the difference in the path traversed by the two waves.
The phase difference and the path difference are directly proportional to each other.
The path difference is that difference that is travelled by the two waves from their respective sources to a give point on the pattern
Complete step by step solution:
The expression for the path difference and the phase difference
$\begin{align}
&Phase\;difference = \dfrac{{2\pi }}{\lambda } \times path\;difference \\
& \Delta \phi = \dfrac{{2\pi }}{\lambda } \times \Delta x \\
\end{align} $
Here, $\Delta \phi $ is the phase difference,
$\Delta x$ Is the path difference
$\lambda $ Is the wavelength
The phase difference between any two particles in a wave determines lack of harmony in the vibrating state of the two particles. i.e., how far one particle leads the other or lags behind the other.
From the relation,
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \Delta x$
$\Delta x = \dfrac{\lambda }{{2\pi }} \times \Delta \phi $ --- (1)
Also, $\lambda = \dfrac{v}{f}$ --- (2)
Here v is the velocity
f is the frequency
We know the value of the frequency and velocity, and now, we calculate the wavelength, from equation (2)
$\begin{align}
&\lambda = \dfrac{{\left( {330m/s} \right)}}{{\left( {50Hz} \right)}} \\
&\Rightarrow \lambda = 6.6m \\
\end{align} $
The wavelength is 6.6m
Now, we have the value of wavelength and the value of phase difference.
We calculate the path difference from an equation (1)
$\begin{align}
&\Delta x = \dfrac{\lambda }{{2\pi }} \times \Delta \phi \\
& \Rightarrow \Delta x = \left( {\dfrac{{6.6}}{{2 \times \pi }}} \right) \times \left( {\dfrac{\pi }{3}} \right) \\
& \Rightarrow \Delta x = \dfrac{{2.2}}{2} \\
& \Rightarrow \Delta x = 1.1m \\
\end{align} $
So, the value of the path difference is 1.1m
So, the option (B) is correct.
Note: The term phase difference, it is the difference in the phase angle of the two waves.
The term path difference, it is the difference in the path traversed by the two waves.
The phase difference and the path difference are directly proportional to each other.
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