Question

The equation of a simple harmonic progressive wave is given by\[y=A\sin (100\pi t-3x)\]. Find the distance between the two particles having a phase difference of \[\dfrac{\pi }{3}\].
A. \[\dfrac{\pi }{9}\]m
B. \[\dfrac{\pi }{18}\]m
C. \[\dfrac{\pi }{6}\]m
D. \[\dfrac{\pi }{3}\]m

Answer 1

Hint: We are provided a simple harmonic progressive wave. While solving such kinds of problems we first of all write the general equation representing the simple harmonic progressive wave and then compare it with the given equation. We need to find the distance between the two particles, that is we need to find the phase difference between the two.

Complete step by step solution:
The general equation of a simple harmonic progressive wave is given by \[y=A\sin (\omega t-kx)\], and the given equation is \[y=A\sin (100\pi t-3x)\]. Comparing the two equations we find that \[k=3\]and \[\omega =100\pi \]
Given the phase difference is \[\dfrac{\pi }{3}\]. We know
 \[k=\dfrac{2\pi }{\lambda }\]
\[ \Rightarrow 3= {\dfrac{2\pi }{\lambda }} \\
 \therefore \lambda = {\dfrac{2\pi }{3}} \]
Also, the relationship between the path difference and phase difference is given by: \[\phi =\dfrac{2\pi }{\lambda }\times p\]
\[\Rightarrow \phi =\dfrac{2\pi }{\lambda }\times p\]
\[ \Rightarrow {\dfrac{\pi }{3}} = {\dfrac{2\pi }{2\pi }} \times 3\times p \\
\Rightarrow p = {\dfrac{\pi }{9}} \]

So, the path difference comes out to be \[\dfrac{\pi }{9}\]m that is the distance between the two particles. So, the correct option is A.

Note: While solving this problem we had taken a leaf out of what we have learnt in Young’s double slit experiment. There, we had found out a relationship between the part difference and the phase difference. We have used the same. This all comes from the interference of light. Interference of light is a phenomenon in which two or more waves interfere with each other and there is redistribution of energy. This is one of the most amazing phenomena to be ever observed and the pattern is worth watching.