Question

The phase difference between two displacements at a certain point $A$ at time ${10}^{-3}s$ apart is $\pi$. Find the distance between point $A$ and $B$, which is ${60}^{\circ }$ out of phase than point $A$, if wave velocity is $x$.

• A. $1.33\times {10}^{3}x$
• B. $0.33\times {10}^{4}x$
• C. $2.33\times {10}^{4}x$
• D. $3.33\times {10}^{4}x$

Answer 1

The correct option is D $3.33\times {10}^{4}x$

We know that:

$\Delta \varphi =2\pi f\left(\Delta t\right)$
Given, time $={10}^{-3}s,$ Phase difference between two displacements $=\pi$
$\Delta \varphi =2\pi f\left(\Delta t\right)$
$\pi =2\pi f\times {10}^{-3}$
$f=500Hz$
Now we have, $\Delta \varphi =k\Delta x,k=\frac{2\pi }{\lambda },v=\lambda f$
The phase difference of the wave is given by
$\Delta \varphi =k\Delta x$
$\Delta x=\frac{\Delta \varphi }{k}=\frac{\left(\Delta \varphi \right)\lambda }{2\pi }=\frac{\left(\Delta \varphi \right)\left(v/f\right)}{2\pi }$
$=\frac{\left(\pi /3\right)\left(x/500\right)}{2\pi }=3.33\times {10}^{-4}x$

Final answer: $\left(d\right)$