Question

Equation of a progressive wave is given by $y=0.2\text{cos}\pi (0.04t+0.02x-\frac{\pi }{6})$
where distances are in $\text{centimeter}$ and time in $\text{seconds}$. What will be the minimum distance between two particles having phase difference of $\frac{\pi }{2}$?

• A. $4\text{cm}$
• B. $8\text{cm}$
• C. $25\text{cm}$
• D. $10\text{cm}$

Answer 1

The correct option is C $25\text{cm}$

Given:
$y=0.2\cos \pi (0.04t+0.02x-\frac{\pi }{6})$
On comparing with standard equation of progressive wave,
$y=A\cos (\omega t+kx+\varphi )$,
Wave number, $k=0.02\pi {\text{cm}}^{-1}$
So, wavelength, $\lambda =\frac{2\pi }{k}=\frac{2\pi }{0.02\pi }=100\text{cm}$
We know phase difference,
$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$
$\Rightarrow \Delta x=\frac{\Delta \varphi \lambda }{2\pi }=\frac{\frac{\pi }{2}\times 100}{2\pi }=25\text{cm}$