If in a stationary wave the amplitude corresponding to antinode is $4 \,cm$, then the amplitude corresponding to a particle of medium located exactly midway between a node and an antinode is ........... $cm$
Medium
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(b)
$y=A_0 \sin (k x) \cos \omega t$
Mid way between a node and antinode is $\frac{\lambda}{8}$ from origin. Function for amplitude is $A=A_0 \sin (k x$
$A=4 \sin \left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}\right)$
$A=2 \sqrt{2} \,cm$
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