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(B) The equation of a stationary wave is given by $y = A_0 \sin(kx) \cos(\omega t)$,where $A_0$ is the amplitude of the antinode.Given,$A_0 = 4 \,cm$.

Vedclass · Accepted Answer

$2 \sqrt{2}$

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(B) The equation of a stationary wave is given by $y = A_0 \sin(kx) \cos(\omega t)$,where $A_0$ is the amplitude of the antinode.Given,$A_0 = 4 \,cm$.The position of a node is where $\sin(kx) = 0$,and the position of an antinode is where $\sin(kx) = 1$.The distance between a node and an adjacent antinode is $\frac{\lambda}{4}$.The particle located midway between a node and an antinode is at a distance $x = \frac{1}{2} 	imes \frac{\lambda}{4} = \frac{\lambda}{8}$ from the node.The amplitude $A$ of a particle at position $x$ is given by $A = A_0 \sin(kx)$.Substituting $k = \frac{2\pi}{\lambda}$ and $x = \frac{\lambda}{8}$:$A = 4 \sin\left(\frac{2\pi}{\lambda} 	imes \frac{\lambda}{8}ight) = 4 \sin\left(\frac{\pi}{4}ight)$.Since $\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$,we get $A = 4 	imes \frac{1}{\sqrt{2}} = 2\sqrt{2} \,cm$.

If in a stationary wave the amplitude corresponding to an antinode is $4 \,cm$,then the amplitude corresponding to a particle of medium located exactly midway between a node and an antinode is ........... $cm$.

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