Question

Consider two identical waves travelling in the positive x-direction. The two waves start at the same position but at different times, t1 and t2. What should the time difference ∆t be in order to have a phase difference of π/4. Assume that the frequency of each wave is f = 20 Hz.

Answer 1

two identical waves travelling in the positive x-direction and start at the same position

$X_1=Asin\phi_1 =Asinw_1t=Asin\frac{2\pi}{T}=Asin2\pi \nu t_1$

$X_2=Asin\phi_2 =Asinw_2t=Asin\frac{2\pi}{T}=Asin2\pi \nu t_2$

$\phi_1-\phi_2 = 2\pi \nu (t_1-t_2)=2\pi \nu \Delta t = \frac{\pi}{4} \to \Delta t = \frac{1}{8\nu} = 6.25 ms$