A wave y = a sin(omega t - kx) on a string me | vedclass

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(B) Let the first wave be $y_1 = a \sin(\omega t - kx)$.Let the unknown wave be $y_2 = A \sin(\omega t + kx + \phi)$.The resultant wave is $y = y_1 +

Vedclass · Accepted Answer

$y = -a \sin(\omega t + kx)$

Vedclass · Answer

(B) Let the first wave be $y_1 = a \sin(\omega t - kx)$.Let the unknown wave be $y_2 = A \sin(\omega t + kx + \phi)$.The resultant wave is $y = y_1 + y_2$.$A$ node is formed at $x = 0$,which means the resultant displacement $y$ must be zero for all $t$ at $x = 0$.$y(0, t) = a \sin(\omega t) + A \sin(\omega t + \phi) = 0$.For this to hold for all $t$,the waves must be equal in magnitude and opposite in phase.Thus,$A = a$ and $\phi = \pi$ (or phase difference of $\pi$).$y_2 = a \sin(\omega t + kx + \pi) = -a \sin(\omega t + kx)$.Therefore,the equation of the unknown wave is $y = -a \sin(\omega t + kx)$.

$A$ wave $y = a \sin(\omega t - kx)$ on a string meets with another wave producing a node at $x = 0$. Then the equation of the unknown wave is

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