# phase difference of two reflected wave

Suppose a tuning fork generates sound waves with a frequency of 100 Hz. The waves travel in opposite directions along a hallway, are reflected by end walls, and return. The hallway is 47.0 m long and the tuning fork is located 14.0 m from one end. What is the phase difference between the two reflected wave at the point of the tuning fork.

My attempt:

For this question, I was able to solve using the path length difference. However I tried to solve it using wave equation,

i.e. $y_1 = A \cos(kx - wt),\ y_2 = A \cos(kx + wt)$

The two waves travel in opposite directions, so $w$ has different signs. I was planning to compute the $kx-w$t term for $y_1$ and $y_2$, then the phase difference will be the difference of those two phases, but I was unable to get the right answer. I think the reflection shift the phases of $y_1$ and $y_2$ by $\pi$, so the effect of reflection can be omitted.

Can anyone please explain what $x$ and $t$ should be when the two reflected waves meet at the point of tuning fork?

Thank you very much for the help.

## 1 Answer

Your path difference method does use the wave equation although you may not realise it.

Suppose you took at "photograph" of the wave. This is called a "wave profile".
You would be observing how y varies with x for a fixed t.

To make things easier assume that you took the photograph at time t = 0 and x = 0 at the tuning fork.
Then compute the values of k|x| at the two walls. Note that it is the magnitude of x that you need. Subtracting those two values will give you the phase difference at reflection and because the waves travel the same distance after reflection to reach the tuning fork, you double that number to get your answer.

The |k| could have been avoided if you used the y = A cos(wt - kx) firm of the wave equation.