Question

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

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

Answer 1

Answer: (D)

At $x=0$ the phase difference should be $\pi$.
Thus, the correct option is (d).
Alternate solution:
$y_2=a\cos \left(\omega t+kx+{\varphi }_0\right)$
$\therefore y=y_1+y_2=a\cos \left(\omega t-kx+\frac{\pi }{3}\right)$
$+a\cos \left(\omega t+kx+{\varphi }_0\right)$
$=2a\cos \left[\omega t+\frac{\frac{\pi }{3}+{\phi }_0}{2}\right]\times \cos \left[kx+\frac{{\phi }_0-\frac{\pi }{3}}{2}\right]$
$y=0$ at $x=0$ for any $t$
$\Rightarrow kx+\frac{{\phi }_0-\frac{\pi }{3}}{2}=\frac{\pi }{2}$ at $x=0$
$\therefore {\varphi }_0=\frac{4\pi }{3}.$
Hence $y_2=a\cos \left(\omega t+kx+\frac{4\pi }{3}\right)$