Question

Two waves passing through a region are represented by
$y=\left(1.0m\right)\sin \left[\right(3.14c{m}^{-1}\times -\left(157s^{-1}\right)t]$
and $y=\left(1.5cm\right)\sin \left[\right(3.14c{m}^{-1}\times -\left(314s^{-1}\right)t]$.
Find the displacement of the particle at $x=4.5cm$ at time $t=5.0ms$.

Answer 1

According to the principle of superposition.

each wave produces its disturbance independent of the other,

and the resultant disturbance is equal to the vector sum of the

individual disturbance.

The displacements of the particle

at x = 4.5 cm

at the time t = 5.0

ms due to the two waves are

$y_1=\left(1.0cm\right)sin\left[\right(3.14c{m}^{-1}\left)\right(4.5cm\left)\right]$

- $\left[\right(157s^{-1}\left)\right(5.0\times {10}^{-3}s\left)\right]$

= (1.0 cm) $sin[4.5\pi -\frac{\pi }{4}]$

= (1,0 cm ) $sin[4\pi +\frac{\pi }{4}]$

= $\frac{1.0cm}{\sqrt{2}}$

and

$y_2=\left(1.5cm\right)$ $sin\left[\right(1.57c{m}^{-1}\left)\right(4.5cm)$

$-\left(314s^{-1}\right)(5.0\times {10}^{-3})]$

= (1.5 cm) sin $[2.25\pi -\frac{\pi }{2}]$

= (1.5 cm) sin $\frac{\pi }{4}$

= - $\frac{1.5cm}{\sqrt{2}}$

The net displacement is

y = $y_1+y_2$

= $\frac{-0.5cm}{\sqrt{2}}$

= - 0.35 cm