Answer:
Wave 1 and Wave 2.
Explanation:
We know that the general equation of a wave is given by :
[tex]y=A\ sin(kx-\omega t)[/tex]
or
[tex]y=A\ cos(kx-\omega t)[/tex]
We know that the speed of a wave is given by :
[tex]v=\dfrac{\omega}{k}[/tex]
Where
[tex]\omega[/tex] = angular speed
k = propagation constant
Wave 1.
[tex]y=0.12\ cos(3x-21t)[/tex]
[tex]v_1=\dfrac{21}{3}=7\ m/s[/tex]
Wave 2.
[tex]y=0.15\ cos(6x+42t)[/tex]
[tex]v_2=\dfrac{42}{6}=7\ m/s[/tex]
Wave 3.
[tex]y=0.13\ cos(6x+21t)[/tex]
[tex]v_3=\dfrac{21}{6}=3.5\ m/s[/tex]
Wave 4.
[tex]y=-0.23\ cos(3x-42t)[/tex]
[tex]v_1=\dfrac{42}{3}=14\ m/s[/tex]
It is clear that wave 1 and 2 have the same speed i.e. 7 m/s. Hence, this is the required solution.
