Calculating path difference

Watch this thread
This discussion is closed.
lolz1411
Badges: 1
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 10 years ago
#1
Two coherent waves have a wavelenth of λ. The phase differen when they meet is 90 degrees (pi/2 radians). What could the path difference be :
(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?

Thanks
0
Drummy
Badges: 10
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 10 years ago
#2
180 would be half a wavelength so 90 =?
0
nevetstreblig
Badges: 7
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 10 years ago
#3
if they arrive 90 degrees out of phase, then (assuming we're talking a normal sin/cosine wave) then one wave is half a wavelength further down the time axis than the other
0
Crowned Temmy
Badges: 7
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 10 years ago
#4
phase difference = 90 degrees= pi/2

from your formula, path difference = ( phase difference * λ) / 2 pi

= (pi/2 * λ ) /2 pi

= λ/4
1
spongpop
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 10 years ago
#5
i dont understand the difference between the path difference and ohase difference can anyone help please asap
0
goodfellow
Badges: 1
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report 10 years ago
#6
(Original post by spongpop)
i dont understand the difference between the path difference and ohase difference can anyone help please asap
Imagine you have two waves, travelling alongside each other. One wave is slightly ahead of the the other. The distance between the two waves is the path difference.

Image

So these two waves have a path difference 'd'.

What is the phase difference?
Well lets look at the equation for a wave first.

y_{(x,t)} = Asin(kx - {\omega}t)

The phase is the term in the brackets, phase = kx - \omegat

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:
y_{top} = Asin(kx - {\omega}t)

the bottom wave will have the equation:
y_{bottom} = Asin(kx - {\omega}t - kd)

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 \frac{\pi}{2}) had a phase difference of -\frac{\pi}{2} it would be a sin graph.
0
goodfellow
Badges: 1
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report 10 years ago
#7
This might also help, derivation of the wave equation:

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 \omega and is given the name 'angular frequency'.
It has the equation \omega = \frac{2\pi}{T}

So if you have a wave, it could have the equation y_{x,t} = Asin[{\omega}(x)].

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 y_{x,t} = Asin[{\omega}(x - vt)].

which some re-arranging:
{\omega}(x - vt) = {\omega}(\frac{x}{v} - t) = \frac{2{\pi}x}{TV} - \frac{2{\pi}t}{T} = \frac{2{\pi}x}{Tf\lambda} - \frac{2{\pi}t}{T} = \frac{2{\pi}x}{\lambda} - \frac{2{\pi}t}{T} = kx - {\omega}t

K is called the wave number.

so the equation of the travelling wave is now:

y_{x,t} = Asin(kx - {\omega}t).
0
Crowned Temmy
Badges: 7
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report 10 years ago
#8
(Original post by spongpop)
i dont understand the difference between the path difference and ohase difference can anyone help please asap

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.
0
spongpop
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report 10 years ago
#9
thnks guys
1
iMeme69
Badges: 1
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10
Report 1 year ago
#10
me neither lmao hope this isnt too late
0
X
new posts
Back
to top
Latest

Were exams easier or harder than you expected?

Easier (37)
25.87%
As I expected (46)
32.17%
Harder (53)
37.06%
Something else (tell us in the thread) (7)
4.9%

Watched Threads

View All