Question

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.

Answer 1

Answer:

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

Explanation:

What matters is the part $kx \pm \omega t$, 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 $\delta t$ ($\delta t >0$), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$.

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (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 $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$, which at the origin is $k \delta x \pm \omega t$. This would mean that, when the original equation has $kx+\omega t$, we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$, and when the original equation has $kx-\omega t$, we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

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