Back
to top
Calculating phase difference
Watch this thread
Announcements
Page 1 of 1
Go to first unread
Skip to page:
Zwitter Ion
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.
#1
How would you work out phase difference between two points given the frequency, distance and wavelength...and how would you express this in radians.
EG:
Wavelength of a wave is 1.2m, Speed is 3.6ms-1,distance between two points P and Q is 0.4m
What is the phase difference....
The mark scheme says 2(pi)/3.....how did they get that??
Any help will be appreciated
Thanks in advance
EG:
Wavelength of a wave is 1.2m, Speed is 3.6ms-1,distance between two points P and Q is 0.4m
What is the phase difference....
The mark scheme says 2(pi)/3.....how did they get that??
Any help will be appreciated
Thanks in advance
0
reply
ukstudent2011
Badges:
0
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
#2
(Original post by Zwitter Ion)
How would you work out phase difference between two points given the frequency, distance and wavelength...and how would you express this in radians.
EG:
Wavelength of a wave is 1.2m, Speed is 3.6ms-1,distance between two points P and Q is 0.4m
What is the phase difference....
The mark scheme says 2(pi)/3.....how did they get that??
Any help will be appreciated
Thanks in advance
How would you work out phase difference between two points given the frequency, distance and wavelength...and how would you express this in radians.
EG:
Wavelength of a wave is 1.2m, Speed is 3.6ms-1,distance between two points P and Q is 0.4m
What is the phase difference....
The mark scheme says 2(pi)/3.....how did they get that??
Any help will be appreciated
Thanks in advance
0
reply
gorilla_baby
Badges:
9
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
#3
phase difference = 2pi x path difference divided by lambda
path difference is like (n + 1/2)lambda etc..
path difference is like (n + 1/2)lambda etc..
0
reply
gorilla_baby
Badges:
9
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
#4
so path difference = 1/3 lambda and thus the two lambda's in the equation cancel out leaving phase difference = 2pi x (1/3) = 2pi/3
that gives you the answer
that gives you the answer
0
reply
Zwitter Ion
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.
#5
So am i right in saying that a general formula to calculate phase difference is:
distance travelled/wavelength x 2(pi) ???
distance travelled/wavelength x 2(pi) ???
0
reply
gorilla_baby
Badges:
9
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
#6
not distance travelled, but the difference in the distance between the two points in terms on lambda, i.e.a multiple or fraction of the wavelength
0
reply
dannyphantom
Badges:
3
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Amirtha
Badges:
4
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
#8
i came across a question with the same information:1. they've given a displacement/distance graph with a straight line running through the x axis saying its a stationary wave at t=0.
2. then they say it has a period of 20ms and wavelength 1.2m. max amplitude is 5mm
they ask what is the phase diff between particles of the string at .4m and .8m.
the MS says 180 degrees/ pi radians
can someone explain this to me?
2. then they say it has a period of 20ms and wavelength 1.2m. max amplitude is 5mm
they ask what is the phase diff between particles of the string at .4m and .8m.
the MS says 180 degrees/ pi radians
can someone explain this to me?
0
reply
astro67
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.
#9
Report
#9
(Original post by Amirtha)
i came across a question with the same information:1. they've given a displacement/distance graph with a straight line running through the x axis saying its a stationary wave at t=0.
2. then they say it has a period of 20ms and wavelength 1.2m. max amplitude is 5mm
they ask what is the phase diff between particles of the string at .4m and .8m.
the MS says 180 degrees/ pi radians
can someone explain this to me?
i came across a question with the same information:1. they've given a displacement/distance graph with a straight line running through the x axis saying its a stationary wave at t=0.
2. then they say it has a period of 20ms and wavelength 1.2m. max amplitude is 5mm
they ask what is the phase diff between particles of the string at .4m and .8m.
the MS says 180 degrees/ pi radians
can someone explain this to me?
0
reply
Amirtha
Badges:
4
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
#10
(Original post by astro67)
Not sure entirely about the nature of the question - straight line displacement as a function of distance doesn't look like a wave but assuming the statement that we're looking at a stationary wave applies, we can draw certain conclusions. A stationary wave with a node at x = 0 and wavelength 1.2m will have nodes at x = 0.6 m, 1.2 m, 1.8 m etc. Points either side of a node will oscillate out of phase with each other, so the phase difference between them will be pi radians or 180 degree. This is true for any points either side of a node. Points at x = 0.4m and x = 0.8m are equal distances from the 0.6 m node so they will oscillate out of phase with each other with equal amplitude to each other.
Not sure entirely about the nature of the question - straight line displacement as a function of distance doesn't look like a wave but assuming the statement that we're looking at a stationary wave applies, we can draw certain conclusions. A stationary wave with a node at x = 0 and wavelength 1.2m will have nodes at x = 0.6 m, 1.2 m, 1.8 m etc. Points either side of a node will oscillate out of phase with each other, so the phase difference between them will be pi radians or 180 degree. This is true for any points either side of a node. Points at x = 0.4m and x = 0.8m are equal distances from the 0.6 m node so they will oscillate out of phase with each other with equal amplitude to each other.
and what if the points are not of equal distance from the node, what will be the phase difference? like between .5m and .8m?
0
reply
astro67
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.
#11
Report
#11
(Original post by Amirtha)
i thought points either side of a node oscillate in phase. how is it not in phase?
and what if the points are not of equal distance from the node, what will be the phase difference? like between .5m and .8m?
i thought points either side of a node oscillate in phase. how is it not in phase?
and what if the points are not of equal distance from the node, what will be the phase difference? like between .5m and .8m?
The motion of a point at distance x along a stationary wave can be described by A*sin(2*pi*x/lambda)*sin(omega*t). For a given value of x (i.e. a fixed point on the string), the motion is sinusoidal with time, with amplitude given by A*sin(2*pi*x/lambda). All the points between consecutive nodes are moving in phase with each other. For 0<x<(lambda/2), the amplitude is positive. For x = (any integer)*(lambda/2), then the amplitude is zero and we have a node. For (lambda/2)<x<lambda, the amplitude is negative and the string travels in the opposite direction, corresponding to a phase shift of pi relative to points on the string either side of the nodes.
0
reply
Ryen
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.
#12
Report
#12
0.4 divided by the wavelength (1.2) so you get the phase difference. To get that answer in radian multiply it by 2 pi.
0
reply
Divya shree
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.
#13
Report
#13
The formula for phase difference = K • delta x Since k = 2 pie / lemda Substituting the value of lemda = 1.2 We get k = 2 pie / 1.2 ,And delta x = 0.4 Hence solving , k • delta x , We get the answer 2 pie / 3
0
reply
shevinBLZ
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.
#14
Report
#14
(Original post by dannyphantom)
Use the formula,
=2 pi d / wavelength
Put the values,
=2(pi)(0.4)/1.2
=2pi/3
Use the formula,
=2 pi d / wavelength
Put the values,
=2(pi)(0.4)/1.2
=2pi/3
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
