Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is described by y2(x, t) = A sin(kx + ωt + φ). The phases (arguments of the sines) are in radians, as usual, and A = 2.1 cm and φ = 0.35π rad.

Choose the answer that correctly describes the waves’ directions of travel.

MultipleChoice :

1) Both waves travel in the negative x direction.

2) Due to the φ term in the phase of wave 2, it does not travel. Wave 1 travels in the negative x-direction.

3) Due to the φ term in the phase of wave 2, it does not travel. Wave 1 travels in the positive x-direction.

4) Both waves travel in the positive x-direction.

5) Wave 1 travels in the negative x-direction, while wave 2 travels in the positive x-direction.

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

7) There is not enough information.

What matters is the part [tex]kx \pm \omega t[/tex], the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time [tex]\delta t[/tex] ([tex]\delta t >0[/tex]), the displacement y of the wave will be determined by the [tex]kx\pm\omega (t+\delta t)[/tex] term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply [tex]\pm \omega (t+\delta t)[/tex].

To know which side, right or left of the origin, would go through the origin after a time [tex]\delta t[/tex] (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation [tex]\delta x[/tex] (we would be looking where in space is what we would have in the future in time). The term would be then [tex]k(x+\delta x)\pm\omega t[/tex], which at the origin is [tex]k \delta x \pm \omega t[/tex]. This would mean that, when the original equation has [tex]kx+\omega t[/tex], we must have that [tex]\delta x>0[/tex] for [tex]k\delta x+\omega t[/tex] to be equal to [tex]kx+\omega\delta t[/tex], and when the original equation has [tex]kx-\omega t[/tex], we must have that [tex]\delta x<0[/tex] for [tex]k\delta x-\omega t[/tex] to be equal to [tex]kx-\omega \delta t[/tex]

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when [tex]kx+\omega t[/tex], the part of the wave on the positive side ([tex]\delta x>0[/tex]) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.