Question

A standing wave pattern is formed on a string. One of the waves is given by equation $y_{1}=a \cos (\omega t-k x+\pi / 3)$ then the equation of the other wave such that at $x=0$ a node is formed :

• A. $y_{2}=a \sin \left(\omega t+ k x+\frac{\pi}{3}\right)$
• B. $y_{2}=a \cos \left(\omega t +k x+\frac{\pi}{3}\right)$
• C. $y_{2}=a \cos \left(\omega t+ k x+\frac{2 \pi}{3}\right)$
• D. $y_{2}-a \cos \left(\omega t+ k x+\frac{4 \pi}{3}\right)$

Answer 1

Answer: (D)

At $x=0$ the phase difference should be $\pi$ for destructive interference
Alternate solution
We use $y_{2}=a \cos \left(\omega t +k x+\phi_{0}\right)$
Resultant after superposition is
$y=y_{1}+y_{2}=a \cos \left(\omega t-k x+\frac{\pi}{3}\right)+a \cos \left(\omega t +k x+\phi_{0}\right)$
$=2 a \cos \left[\omega t+\frac{\frac{\pi}{3}+\phi_{0}}{2}\right] \times \cos \left[k x+\frac{\phi_{0}-\frac{\pi}{3}}{2}\right]$
$\Rightarrow y=0$ at $x=0$ for any $t$
$\Rightarrow k x+\frac{\phi_{0}-\frac{\pi}{3}}{2}=\frac{\pi}{2}$ at $x=0$
$\Rightarrow \phi_{0}=\frac{4 \pi}{3} .$
Hence $y_{2}=a \cos \left(\omega t +k x+\frac{4 \pi}{3}\right)$