Calculating path differenceWatch this thread
(A) 2λ (B)λ (C) λ/2 (D) λ/4
I understand the formula for phase difference is:
Phase difference= 2 x pi x path difference/ λ
But seeing as none of the answers have pi in them, how am I meant to work this one out?
from your formula, path difference = ( phase difference * λ) / 2 pi
= (pi/2 * λ ) /2 pi
So these two waves have a path difference 'd'.
What is the phase difference?
Well lets look at the equation for a wave first.
The phase is the term in the brackets, phase = kx - t
So look at the bottom wave in the picture, it is shifted to the right by an amount. It travels farther by the path d.
So the top wave will have the equation:
the bottom wave will have the equation:
Then from comparing the two, we can see the 'phase difference' is "kd".
(Phase top) - (Phase bottom) = kd
So if a cos graph (which is behind a sin graph by an amount ) had a phase difference of it would be a sin graph.
Recap on graph transformations:
(1) y = Af(x) stretches the graph of y = f(x) vertically by an amount A (multiply y-coords by A. This constant is called the amplitude.
(2) y = f(x - B) shifts the graph of y = f(x) to the right by an amount B.
(3) y = f[G(x)] squashes the graph of y = f(x) by an amount 1/G (Multiply x-coords by 1/G).
So if you take a sin graph and squash it by a factor of 99, there will be 99x as many peaks/troughs in a certain time. So its frequency has increased. (frequeny = number of peaks that pass a given point in a time) G has the unit and is given the name 'angular frequency'.
It has the equation
So if you have a wave, it could have the equation .
If it is moving to the right at a speed v, in a time 't', it will have moved a distance 'vt'
So the equation becomes .
which some re-arranging:
K is called the wave number.
so the equation of the travelling wave is now:
Path difference is the fraction of wavelength by which one wave leads or lags behind another.
Phase difference is the angle (in radians) by which one wave leads or lags behind another.