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

Answer

Correct option: B.
$2 \sqrt{2}$
b
(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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