A standing wave pattern is formed on a string | vedclass

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(D) The resultant wave is $y = y_1 + y_2$. Let $y_2 = a \cos(\omega t + kx + \phi_0)$.Using the superposition principle,$y = a [\cos(\omega t - kx + \

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$y_2 = a \cos(\omega t + kx + 4\pi/3)$

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(D) The resultant wave is $y = y_1 + y_2$. Let $y_2 = a \cos(\omega t + kx + \phi_0)$.Using the superposition principle,$y = a [\cos(\omega t - kx + \pi/3) + \cos(\omega t + kx + \phi_0)]$.Using the identity $\cos A + \cos B = 2 \cos((A+B)/2) \cos((A-B)/2)$:$y = 2a \cos(\omega t + (kx + \phi_0 - kx + \pi/3)/2) \cos((-kx + \pi/3 - kx - \phi_0)/2)$.$y = 2a \cos(\omega t + (kx + \phi_0)/2 + \pi/6) \cos(-kx + (\pi/3 - \phi_0)/2)$.For a node at $x = 0$,the amplitude must be zero for all $t$,which means the spatial part must be zero at $x = 0$.$\cos(0 + (\pi/3 - \phi_0)/2) = 0$.This implies $(\pi/3 - \phi_0)/2 = \pi/2$ or $3\pi/2$.For $(\pi/3 - \phi_0)/2 = \pi/2$,we get $\pi/3 - \phi_0 = \pi$,so $\phi_0 = -2\pi/3$,which is equivalent to $4\pi/3$ (since $4\pi/3 - 2\pi = -2\pi/3$).Thus,$y_2 = a \cos(\omega t + kx + 4\pi/3)$.

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

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