Question

Equations of a stationary wave and a travelling wave are $y_1=a\sin kx\cos \omega t$ and $y_2=a\sin (\omega t-kx)$. The phase difference between two points $x_1=\frac{\pi }{3k}$ and $x_2=\frac{3\pi }{2k}$ is $\varphi$ for the first wave and ${\varphi }_2$ for the second wave. Find the ratio $\frac{{\varphi }_1}{{\varphi }_2}$.

Answer 1

Phase difference between two points in a standing wave $=n\pi$

Where $n=$ number of nodes between two points

given points are $x_1=\frac{\pi }{3K}{x}_2=\frac{3\pi }{2K}$

Equation of standing wave

$y_1=a\sin kx\cos wt$

at node positions $kx=n\pi$

$x=\frac{n\pi }{K}$

where $n=0,1,2,\mathrm{3...}$

So nodes are at $x=\frac{\pi }{K},\frac{2\pi }{K},\frac{3\pi }{K},...$

So there is only one node between ${ x }_{ 1 }$and${ x }_{ 2 }$hencephasedifference${ \phi }_{ 1 }=\pi$

For the travelling wave

${\varphi }_2=\frac{2\pi }{\lambda }\Delta x$

For equation

$y_2=a\sin kx\cos wt$

So $K=\frac{2\pi }{\lambda }\therefore {\varphi }_2=K[x_2-x_1]=K[\frac{3\pi }{2K}-\frac{\pi }{3K}]=\pi [\frac{3}{2}-\frac{1}{3}]=\frac{7}{6}\pi \therefore \frac{{\varphi }_1}{{\varphi }_2}=\frac{\pi }{\frac{7}{6}\pi }=\frac{6}{7}$