Question

Equations of a stationary wave and a travelling wave are $y_1=1sin\left(kx\right)cos\left(\omega t\right)$ and $y_2=asin(\omega t-kx)$.The phase difference between two points $x_1=\frac{\pi }{3k}$ and $x_2=\frac{3\pi }{2k}$ is ${\varphi }_1$ for the first wave and ${\varphi }_2$ for the second wave.The ratio $\frac{{\varphi }_1}{{\varphi }_2}$ is

• A. 1
• B. 5/6
• C. 3/4
• D. 6/7

Answer 1

The correct option is D 6/7

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

Where n is number of nodes between two points.

Given points are $x_1=\frac{\pi }{3k}=\frac{60}{k}$

$x_2=\frac{3\pi }{2k}=\frac{210}{k}$

Equation of the standing wave

$y_1=a\sin kx\cos \omega t$

At node points $kx=n\pi$

$x=\frac{n\pi }{k}(n=0,1,2,\mathrm{3...})$

So nodes are =$\frac{\pi }{k},\frac{2\pi }{k}....$

$=\frac{180}{k},\frac{360}{k}....$

Since there is only one node between phase difference ${\varphi }_1=\pi$

For travelling wave ${\varphi }_2=\frac{2\pi }{\lambda }△x$

From the equation

$y_2=a\sin (\omega t-kx)$

$k=\frac{2\pi }{\lambda }$

$\therefore {\varphi }_2=k[x_2-x_1]=k[\frac{3\pi }{2k}-\frac{\pi }{3k}]=\frac{7}{6}\pi$

$\therefore \frac{{\varphi }_1}{{\varphi }_2}=\frac{\pi }{\frac{7}{6}\pi }=\frac{6}{7}$