Question

The equation of a stationary and a travelling waves are $y_1=a\sin kx\cos \omega t$ and $y_2=a\sin (\omega t-kx)$ respectively. The phase difference between two points, $x_1=\frac{\pi }{3k}$ and $x_2=\frac{3\pi }{2k}$ is ${\varphi }_1$ in the standing wave $y_1$ and is ${\varphi }_2$ in travelling wave $y_2$ then the ratio $\frac{{\varphi }_1}{{\varphi }_2}$ is

• A. $\frac{1}{3}$
• B. $\frac{5}{6}$
• C. $\frac{3}{4}$
• D. $\frac{6}{7}$

Answer 1

The correct option is D $\frac{6}{7}$

For the stationary wave,
$y_1=a\sin kx\cos \omega t$
Wave number,
$k=\frac{2\pi }{\lambda }$
Hence, two points on standing wave are,
$x_1=\frac{\pi }{3k}=\frac{\lambda }{6}$, and
$x_2=\frac{3\pi }{2k}=\frac{3\lambda }{4}$


It is clear from the above figure that points $x_1$ and $x_2$ lie in the adjacent loops.
So, the phase difference between them will be $\pi$.
$\Rightarrow {\varphi }_1=\pi$ ..............(1)

For the travelling wave,
$y_2=a\sin (\omega t-kx)$
Wave number,
$k=\frac{2\pi }{\lambda }$
Hence, two points on the travelling wave are,
$x_1=\frac{\pi }{3k}=\frac{\lambda }{6}$, and
$x_2=\frac{3\pi }{2k}=\frac{3\lambda }{4}$
$\Delta x=x_2-x_1=\frac{3\lambda }{4}-\frac{\lambda }{6}=\frac{7\lambda }{12}$
Phase difference,
${\varphi }_2=\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \frac{7\lambda }{12}$
$=\frac{7\pi }{6}$ .............(2)
Now,
$\frac{{\varphi }_1}{{\varphi }_2}=\frac{\pi }{\frac{7\pi }{6}}$ [from (1) and (2)]
$\Rightarrow \frac{{\varphi }_1}{{\varphi }_2}=\frac{6}{7}$