Question

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

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

Answer 1

Answer: (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 $A=A_{0} \sin (k x)$
$A=4 \sin \left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}\right)$
$A=2 \sqrt{2} cm$