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(C) The equation of a standing wave is given by $y(x, t) = A_m \sin(kx) \cos(\omega t)$,where $A_m$ is the maximum displacement amplitude (antinode am

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$10$

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(C) The equation of a standing wave is given by $y(x, t) = A_m \sin(kx) \cos(\omega t)$,where $A_m$ is the maximum displacement amplitude (antinode amplitude).The displacement amplitude at a position $x$ is given by $A(x) = |A_m \sin(kx)|$.Given $A(10) = 5\sqrt{3} \ mm$,we have $5\sqrt{3} = A_m \sin(10k)$.It is given that the distance between two points in the same loop having the same amplitude $5\sqrt{3} \ mm$ is $10 \ cm$. Let the two points be $x_1$ and $x_2$. Since they are in the same loop,$x_2 - x_1 = 10 \ cm$.Due to the symmetry of the sine function about the antinode,if the antinode is at $x_0$,then $x_1 = x_0 - d$ and $x_2 = x_0 + d$. Thus,$x_2 - x_1 = 2d = 10 \ cm$,so $d = 5 \ cm$.At $x_1 = 10 \ cm$,we have $A_m \sin(10k) = 5\sqrt{3}$.At $x_2 = 20 \ cm$,we have $A_m \sin(20k) = 5\sqrt{3}$.Since $\sin(10k) = \sin(20k)$,we must have $10k = \pi - 20k$ (as they are in the same loop).$30k = \pi \implies k = \frac{\pi}{30} \ cm^{-1}$.Substituting $k$ back into the amplitude equation:$5\sqrt{3} = A_m \sin\left(\frac{\pi}{30} 	imes 10ight) = A_m \sin\left(\frac{\pi}{3}ight) = A_m \left(\frac{\sqrt{3}}{2}ight)$.$A_m = 5\sqrt{3} 	imes \frac{2}{\sqrt{3}} = 10 \ mm$.

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