Question

The period of oscillations of a point is 0.04 sec and the velocity of propagation of oscillation is 300 m/sec. The difference of phases between the oscillations of two points at distances 10 and 16 m respectively from the source of oscillation is:
(A) $2\pi $
(B) $\dfrac{\pi }{2}$
(C) $\dfrac{\pi }{4}$
(D) $\pi $

Answer 1

Hint: To answer this question we have to find the relation between the path difference and the phase difference in the beginning. Then we have to find the phase difference at 10 m from the source of oscillation. After that the same we have to find for 16 m. At the end we have to find the value for the phase difference between 16 m and 10 m from the source. This will give the answer to the required question.

Complete step by step answer:
We should know that the time period of the wave is given as t. The value of t is 0.04 seconds.
Now the velocity of propagation of oscillation is given as v. The value of v is 300 m/ sec.
So now the wavelength of the oscillation is given as $\lambda $. The value of $\lambda $ is given as $\lambda = v \times t$.
Now we have to substitute the values from the question to the given expressions.
So the value of wavelength or $\lambda $becomes :
$
  \lambda = 300 \times 0.04 \\
   \Rightarrow \lambda = 12m \\
 $
Now we have to use the relation between the path difference that is $\Delta x$ and the phase difference, which is given below as:
$\dfrac{{\Delta x}}{\lambda } = \dfrac{{\Delta \phi }}{{2\pi }}$
Now it is known that we have to find the phase difference at 10 m from the source of oscillation. So given below are the expressions:
$\Delta x = 10$
So we can write that:
$
  \dfrac{{10}}{{12}} = \dfrac{{\Delta \phi }}{{2\pi }} \\
   \Rightarrow \dfrac{{10}}{{12}} \times 2\pi = \Delta \phi \\
   \Rightarrow \Delta \phi = \dfrac{{5\pi }}{3} \\
 $
Now we have to find the phase difference at 16 m from the source of the oscillation. So the expressions are as follows:
$\Delta x = 16$
So we can write that:
$
  \dfrac{{16}}{{12}} = \dfrac{{\Delta \phi }}{{2\pi }} \\
   \Rightarrow \dfrac{{16}}{{12}} \times 2\pi = \Delta \phi \\
   \Rightarrow \Delta \varphi = \dfrac{{8\pi }}{3} \\
 $
Now we have to find the phase difference at 16 m and 10 m from the source. So the expressions are as follows:
$
  \Delta {\phi _{16}} - \Delta {\phi _{10}} = \dfrac{{8\pi }}{3} - \dfrac{{5\pi }}{3} \\
   \Rightarrow \Delta {\phi _{16}} - \Delta {\phi _{10}} = \dfrac{{3\pi }}{3} \\
   \Rightarrow \Delta {\phi _{16}} - \Delta {\phi _{10}} = \pi \\
 $
So we can say that the difference of phases between the oscillations of two points at distances 10 and 16 m respectively from the source of oscillation is $\pi $.

Hence the correct answer is option D.

Note: We should know that oscillation is defined as the repetitive variation with respect to time of some specific measure which is about a central value, about the point of equilibrium. The state of oscillation may vary between two different states.
We have also come across the term wavelength. By wavelength we mean the length of one wave or we can say the distance from a specific point on one wave to the same point in case of another wave.