Question

Two waves passing through a region are represented by

$y=1.0cmsin\left[\right(3.14c{m}^{-1})x-(157s^{-1}\left)t\right]$

and $y=1.5cmsin\left[\right(1.57c{m}^{-1})x-(314s^{-1}\left)t\right]$

Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.

• A. 0.35 cm
• B.
• C.
• D. - 0.35 cm

Answer 1

The correct option is D

- 0.35 cm

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 disturbances. The displacements of the particle at x = 4.5 cm at time t = 5.0 ms due to the two waves are,

$y_1=\left(1.0cm\right)sin\left(3.14c{m}^{-1}\right)\left(4.5cm\right)-\left(157s^{-1}\right)(5.0\times {10}^{-3}s)$

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

$=\left(1.0cm\right)sin[4.5\pi -\frac{\pi }{4}]=\frac{1.0cm}{\sqrt{2}}$

and

$y_2=\left(1.5cm\right)sin\left[\right(1.57c{m}^{-1}\left)\right(4.5cm)-(314s^{-1}\left)\right(5.0\times {10}^{-3}s\left)\right]$

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

= $\left(1.5cm\right)sin[2\pi -\frac{\pi }{4}]$

= $-\left(1.5cm\right)sin\frac{\pi }{4}=-\frac{1.5cm}{\sqrt{2}}$

The net displacment is

$y=y_1+y_2=-\frac{0.5cm}{\sqrt{2}}=-0.35cm$