Question

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

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

Answer 1

The correct option is B $2\sqrt{2}\text{cm}$

Equation of a stationary wave is given by
$y=A_0\sin \left(kx\right)\cos \left(\omega t\right)....\left(1\right)$

Mid way between a node and antinode is $\frac{\lambda }{8}$ from the origin.

From $\left(1\right)$,
Function for amplitude is $A=A_0\sin \left(kx\right)$
$A=4\sin (\frac{2\pi }{\lambda }\times \frac{\lambda }{8})$
$A=2\sqrt{2}\text{cm}$