Question

A standing wave pattern is formed on a string. One of the waves is given by equation $y_1=a\cos (\omega t-kx+\frac{\pi }{3})$, then the equation of the other wave, such that at $x=0$ a node is formed.

• A. $y_2=a\cos (\omega t+kx+\frac{\pi }{3})$
• B. $y_2=a\cos (\omega t+kx+\frac{4\pi }{3})$
• C. $y_2=a\sin (\omega t+kx+\frac{\pi }{3})$
• D. $y_2=a\cos (\omega t+kx+\frac{2\pi }{3})$

Answer 1

The correct option is B $y_2=a\cos (\omega t+kx+\frac{4\pi }{3})$

At $x=0$ the phase difference should be $\pi$.
$\therefore$ the correct option is $\left(d\right)$
Alternate solution
Given
$y_1=a\cos (\omega t-kx+\frac{\pi }{3})$

Let,
$y_2=a\cos (\omega t+kx+{\varphi }_0)$
$\therefore y=y_1+y_2$
$y=a\cos (\omega t-kx+\frac{\pi }{3})+y_2=a\cos (\omega t+kx+{\varphi }_0)$
$y=2a\cos [\omega t+\frac{\frac{\omega }{3}+{\varphi }_0}{2}]\times \cos [kx+\frac{{\varphi }_0-\frac{\pi }{3}}{2}]$
$\therefore y=0$ at $x=0$ for any $t$
then,
$\cos [kx+\frac{{\varphi }_0-\frac{\pi }{3}}{2}]=0$

$kx+\frac{{\varphi }_0-\frac{\pi }{3}}{2}=\frac{\pi }{2}$
Hence,
$y_2=a\cos (\omega t+kx+\frac{4\pi }{3})$

Final answer: (d)