Question

Four waves are described by the following equations, where distances are measured in meters and times in seconds. I. y = 0.12 cos(3x - 21t) II. y = 0.15 sin(6x + 42t) III. y = 0.13 cos(6x + 21t) IV. y = -0.23 sin(3x - 42t) Which of these waves have the same speed?

Answer 1

Answer:

Wave 1 and Wave 2.      

Explanation:

We know that the general equation of a wave is given by :

$y=A\ sin(kx-\omega t)$    

or

$y=A\ cos(kx-\omega t)$    

We know that the speed of a wave is given by :

$v=\dfrac{\omega}{k}$

Where

$\omega$ = angular speed

k = propagation constant

Wave 1.

$y=0.12\ cos(3x-21t)$

$v_1=\dfrac{21}{3}=7\ m/s$

Wave 2.

$y=0.15\ cos(6x+42t)$

$v_2=\dfrac{42}{6}=7\ m/s$

Wave 3.

$y=0.13\ cos(6x+21t)$

$v_3=\dfrac{21}{6}=3.5\ m/s$

Wave 4.

$y=-0.23\ cos(3x-42t)$

$v_1=\dfrac{42}{3}=14\ m/s$

It is clear that wave 1 and 2 have the same speed i.e. 7 m/s. Hence, this is the required solution.