Question

Two waves passing through a region are represented by $y_1=(2.0cm)\sin \left[\left(3.14c{m}^{-1}\right)x-\left(157s^{-1}\right)t\right]$ and $y_2=(2.5cm)\sin \left[\left(1.57c{m}^{-1}\right)x-\left(314s^{-1}\right)t\right]$ The magnitude displacement of the particle at $x=4.5cm$ at time $t=5.0ms$ is__________ $cm$

Answer 1

Answer: 0.35

$y_1=(2.0cm)\sin [\left(3.14c{m}^{-1}\right)(4.5cm)$
$-\left(157s^{-1}\right)\left(5.0\times 10^{-3}s\right)]$
$=(2.0cm)\sin \left[4.5\pi -\frac{\pi }{2}\times 100\times 5.0\times 10^{-3}\right]$
$=(2.0cm)\sin \left[4.5\pi -\frac{\pi }{4}\right]$
$=(2.0cm)\sin \left(4\pi +\frac{\pi }{4}\right)=\frac{2.0cm}{\sqrt{2}}$
$y_2=(2.5cm)\sin [\left(1.57c{m}^{-1}\right)(4.5cm)$
$-\left(314s^{-1}\right)\left(5.0\times 10^{-3}\right)]$
$=(2.5cm)\sin \left[2.25\pi -\frac{\pi }{2}\right]$
$=(2.5cm)\sin \left[2\pi -\frac{\pi }{4}\right]$
$=-\frac{2.5cm}{\sqrt{2}}$
Net displacement:
$y=y_1+y_2=-\frac{0.5}{\sqrt{2}}=-(0.5)\times 0.70=-0.35cm$