Question

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

• A. $2cm$
• B. $2\sqrt{2}cm$
• C. $\sqrt{2}cm$
• D. $1.5cm$

Answer 1

Answer: (B)

$y=A_0\sin (kx)\cos \omega t$
Mid way between a node and antinode is $\frac{\lambda }{8}$ from origin.
Function for amplitude is $A=A_0\sin (kx)$
$A=4\sin \left(\frac{2\pi }{\lambda }\times \frac{\lambda }{8}\right)$
$A=2\sqrt{2}cm$