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

Wave 1 and Wave 2.

Step-by-step 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=(\omega)/(k)`

Where

`\omega` = angular speed

k = propagation constant

Wave 1.

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

`v_1=(21)/(3)=7\ m/s`

Wave 2.

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

`v_2=(42)/(6)=7\ m/s`

Wave 3.

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

`v_3=(21)/(6)=3.5\ m/s`

Wave 4.

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

`v_1=(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.