A standing wave pattern is formed on a string. One of the waves i… - Vidyadip

MCQ

A standing wave pattern is formed on a string. One of the waves is given by equation ${y_1} = a\,\cos \,\left( {\omega t - kx + \pi /3} \right)$ 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
${y_2} = a\,\cos \,\left( {\omega t + kx + \frac{{2\pi }}{3}} \right)$
✓
${y_2} = a\,\cos \,\left( {\omega t + kx + \frac{{4\pi }}{3}} \right)$

✓

Answer

Correct option: D.

${y_2} = a\,\cos \,\left( {\omega t + kx + \frac{{4\pi }}{3}} \right)$

d
At $x\ =\ 0$ the phase difference should be $\pi $
$\therefore$ the correct option is $D$

${y_2} - a\,\cos \left( {\omega t + kx + {\phi _0}} \right)$

$\therefore y = {y_1} + {y_2} = a\,\cos \left( {\omega t - kx + \frac{\pi }{3}} \right)$

$ + a\,\cos \left( {\omega t + kx + {\phi _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]$

$\because $ $y = 0$ at $x = 0$ for any $t$

$ \Rightarrow kx + \frac{{{\phi _0} - \frac{\pi }{3}}}{2} = \frac{\pi }{2}$

$\therefore {\phi _0} = \frac{{4\pi }}{3}.$ Hence ${y_2} = a\,\cos \left( {\omega t + kx + \frac{{4\pi }}{3}} \right)$

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