Question

How is the path length difference calculated in objective question 1 of chapter 18 (Physics for Scientists and Engineers, ninth edition)? Why is it changed by 0.2 m upon sliding the tube up 0.1m?

Answer 1

Ans:

Phase constant = 90⁰

A (Amplitude) = 0.04 m

Use the general expression for the wave function for a sinusoidal wave traveling to the right is:

Y = Asin(kx – wt + φ)

Each wave has the same amplitude A, and second wave is 90⁰ out of phase:

Y1 = A sin (kx-wt)

Y2 = A sin (kx – wt + 90)

Using principle of superposition

Y = Y1 + Y2

            = A sin (kx-wt) + A sin (kx – wt + 90)

            = A [sin (kx-wt) + sin (kx-wt+90)]

Use following trigonometric identity:

sin a + sin b = 2 sin ((a+b)/2) cos ((a-b)/2)

So,

Y = 2A sin [(sin (kx-wt) + sin (kx-wt+90))/2] cos [(sin (kx-wt) - sin (kx-wt+90))/2]

            = 2A sin [(2kx – 2wt – 90)/2] cos [90/2]

            = 2A sin (kx – wt – 45) cos 45

Here,

Amplitude A’ = 2A cos 45 = 0.0566m