Question

The distance between two points differing in phase by ${60}^0$ on a wave having a wave velocity $360m{s}^{-1}$ and frequency $500Hz$ is
(A) $0.72m$
(B) $0.18m$
(C) $0.12m$
(D) $0.32m$

Answer 1

Hint: In order to solve this question, we will first calculate the wavelength of the wave and then by converting the given phase difference in radians and then using the formula for phase difference and path difference we will solve for path difference.

Formula Used:
$$\lambda ={\displaystyle \frac{v_{velocity}}{{\nu }_{frequency}}}$$
where, $\lambda$ is the wavelength defined as the ratio of velocity and frequency of the wave.
Path difference is related to phase difference as,
$\mathrm{\Delta }x={\displaystyle \frac{\lambda }{2\pi }}\mathrm{\Delta }\varphi$

Complete step by step solution:
We have given the values of following parameters as,
$$\begin{gathered} \mathrm{\Delta }\varphi ={60}^0={\displaystyle \frac{\pi }{3}} \\ \Rightarrow {\nu }_{\text{frequency}}=500Hz \\ \Rightarrow v_{velocity}=360m{s}^{-1} \end{gathered}$$
Now, using the formula $\lambda ={\displaystyle \frac{v_{velocity}}{{\nu }_{frequency}}}$ we get the wavelength of the wave as
$$\begin{gathered} \lambda ={\displaystyle \frac{360}{500}} \\ \Rightarrow \lambda =0.72m \end{gathered}$$
Now, again using the formula $\mathrm{\Delta }x={\displaystyle \frac{\lambda }{2\pi }}\mathrm{\Delta }\varphi$ we get the separation between two points also called path difference as
$$\begin{gathered} \mathrm{\Delta }x={\displaystyle \frac{0.72\times \pi }{2\pi \times 3}} \\ \therefore \mathrm{\Delta }x=0.12m \end{gathered}$$
So, The separation between two points having a phase difference of ${60}^0$ is $0.12m$.

Hence, the correct option is (C).

Note: It should be remembered that always convert the given angle of phase difference from degree to radians and the general conversion relation is $1^0={\displaystyle \frac{\pi }{180}}radians$ and always ensure the units of frequency, velocity, and wavelength in same standards.