Question

A wave of frequency $680\text{Hz}$ has a wave velocity of $340\text{m/s}$ . Find the distance between two points on a wave whose phase difference is ${45}^{\circ }.$

• A. $0.625\text{cm}$
• B. $0.625\text{m}$
• C. $6.25\text{cm}$
• D. $6.25\text{m}$

Answer 1

The correct option is C $6.25\text{cm}$

We know that ,
phase difference $\left(\Delta \varphi \right)=\frac{2\pi }{\lambda }\times$ path difference $\left(\Delta x\right)$

$\Rightarrow \Delta x=\frac{\lambda }{2\pi }\left(\Delta \varphi \right)$

But, $v=f\lambda$ using this in the above equation we get ,

$\Delta x=\frac{v}{2\pi f}\left(\Delta \varphi \right)$
From the data given in the question,
$\Delta x=\frac{340}{2\times \pi \times 680}\times \frac{\pi }{4}=\frac{1}{16}=0.0625\text{m}\text{or}6.25\text{cm}$

Thus, option (c) is the correct answer.