(4-3)   ①  Amplitude (maximum displacement) = 10

Velocity (distance moved by peak / time taken)  =  2.5 / 2  =  1.25. 

Wavelength (shortest distance between 2 points that are in phase) = 5

When x and t are 0 (top picture), y=0.

②  Substituting into IV-1, we get: y = 10sin(2π(x-1.25t)/5)   =   10sin(0.4π(x-1.25t))

③  Since the wavelengths and velocities are the same, the two waves keep the same separation – that’s to say, they maintain the same phase relationship, as the waves travel. So all we need to do is to add a phase correction, just like we did in Question 4-1, part (d). In this case, you can see that at any point on the orange wave, we need to add ⅝ to the value of x, in order to get the correct value of y, using the equation from part ②. 

So - following the example of Question 4-1, we could write the equation  for the phase-shifted wave as:

y = 10sin(2π(x-1.25t + ⅝)/5)

Since the additional term is a constant, however, independent of x and t, we can separate it out as (2π x ⅝)/5 = π/4. 

So we can just add this phase correction to adjust the equation to describe the orange wave:

                                                  y = 10sin(2π(x-1.25t)/5  + π/4)

Notice that π/4 is equivalent to ⅛ of a wavelength, if the wavelength is (as in an unscaled sine curve) 2π. This is the way these phase shifts are usually represented.

④   The generalisation of this, then, just requires us to add in the phase shift (P), expressed relative to 2π = 1 wavelength:

If the wave is shifted by Δx, we can see - following the example of ③ - that P will just be given by

P = 2π x Δx/λ

y = Asin(2π(x-vt)/λ + P)