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phase difference
![]([url="http://img76.imageshack.us/my.php?image=phasevq8.gif"][img]http://img76.imageshack.us/img76/1999/phasevq8.th.gif[/img][/url])
A second transverse wave T2,of amplitude A has the same waveform as wave t1 but lags behind T1 by a phase angle of 60 degree.The two waves T1 and T2 pass through the same point.
On fig.5.1,draw the variation with time t of the displacement x of the point in wave T2.
Please explain me this thoroughly as it has always posed me a problem.If you have enough time,please explain me about the phase difference as well.
The question just says draw T2 60 degrees out of phase with T1. And it "lags behind", so you draw it to the left of T1.
A full wave cycle is 360 degrees
A full wave cycle is 360 degrees
Say, for exmaple, that we have and plotted on the same axis. The latter of the two graphs is the same as the first, but shifted to the right; we thus say it is out-of-phase by 60 degrees, or
In the example that you gave, T2 is, as above, 60 degrees lagging out of phase with T1; you should therefore plot the max, min and zero x-value points of T1 60 degrees to the right, and join them up.
In the example that you gave, T2 is, as above, 60 degrees lagging out of phase with T1; you should therefore plot the max, min and zero x-value points of T1 60 degrees to the right, and join them up.
How to do calculations for this?Can you show it to me?
I mean how many steps i have to take in y-axis and x-axis to draw T2
Ok, well....
The question says that T2 has the same amplitude as T1; T2 therefore peaks at A on the y-axis the same as T1.
Secondly, we can see from the graph that one complete wave (of T1) takes 3 seconds, and therefore
If - that is, the time at which the new wave is 60 degrees out-of-phase with T1. The question states that it is LAGGING from T1, and thus we need to plot T2 0.5s behind T1.
EDIT: That is, to the RIGHT
of T1....
The question says that T2 has the same amplitude as T1; T2 therefore peaks at A on the y-axis the same as T1.
Secondly, we can see from the graph that one complete wave (of T1) takes 3 seconds, and therefore
If - that is, the time at which the new wave is 60 degrees out-of-phase with T1. The question states that it is LAGGING from T1, and thus we need to plot T2 0.5s behind T1.
EDIT: That is, to the RIGHT
of T1....Thanks a lot pal but i still cannot understand plotting on the y-axis.Can you help me through this.On y-axis can you use the same formula for phase difference.
There is no 'set formula' for phase difference, as far as I am aware...
Also, I'm still unsure as to why you're confused with plotting on the vertical axis; there is nothing different to plot; the waves amplitude is the same. T2 is exactly the same as T1, but every point on the wave has just shifted left by 5 little squares (0.5 seconds) - the working for that is above. If you're still stuck, can you be a little more concise as to what you're exactly having difficulty with...?
Also, I'm still unsure as to why you're confused with plotting on the vertical axis; there is nothing different to plot; the waves amplitude is the same. T2 is exactly the same as T1, but every point on the wave has just shifted left by 5 little squares (0.5 seconds) - the working for that is above. If you're still stuck, can you be a little more concise as to what you're exactly having difficulty with...?
Mitch87
The question states that it is LAGGING from T1, and thus we need to plot T2 0.5s behind T1.
I have attached a (shoddy) sketch of what T2 should look like.
I have attached a (shoddy) sketch of what T2 should look like.
Your sketch is wrong.
By plotting "T2 0.5s behind T1", you mean drawing the T2 waveform to the RIGHT of T1, since this is later in time.
If T2 lags T1, then this means T2 reaches its peak 0.5s after T1 i.e. if T1's peak is at 3s, then T2's must be at 3.5s.
roshanhero
Thanks a lot pal but i still cannot understand plotting on the y-axis.Can you help me through this.On y-axis can you use the same formula for phase difference.
The y-axis in the case is the displacement of the medium at a single point in space over time. Imagine looking at a rubber duck bobbing up and down in the water as shallow waves pass it. This is (almost) looking at a single point in space as the wave passes. If you plotted the displacement of the duck (y) from its equilibrium position (zero) as a function of time (t), you would put y on the vertical axis and t on the horizontal axis. This is the diagram we have.
Note that it is the displacement at a single point in space, as the waves pass.
Ok worzo can you help me to draw wave T2 then.
Worzo
Your sketch is wrong.
By plotting "T2 0.5s behind T1", you mean drawing the T2 waveform to the RIGHT of T1, since this is later in time.
If T2 lags T1, then this means T2 reaches its peak 0.5s after T1 i.e. if T1's peak is at 3s, then T2's must be at 3.5s.
By plotting "T2 0.5s behind T1", you mean drawing the T2 waveform to the RIGHT of T1, since this is later in time.
If T2 lags T1, then this means T2 reaches its peak 0.5s after T1 i.e. if T1's peak is at 3s, then T2's must be at 3.5s.
If its to the right of T1 though, this implies it has been in existence longer. When the question says "0.5secs behind, doesn't it mean it was initiated 0.5 secs after T1??
roshanhero
Ok worzo can you help me to draw wave T2 then.
You draw exactly the same shape because the amplitude is the same.
Because it is 60 degrees out of phase, and one full wave cycle is 360 degrees...... you draw it 0.5 secs behind because 0.5 secs is equal to 60/360 of one full cycle. 60/360 is the same as 10/60 which is the same as 1/6. So you draw the wave 1/6th of a cycle back.
PHKnows
If its to the right of T1 though, this implies it has been in existence longer. When the question says "0.5secs behind, doesn't it mean it was initiated 0.5 secs after T1??
Your first sentence is wrong, your second sentence is correct.
PHKnows
You draw exactly the same shape because the amplitude is the same.
...and it has the same frequency.
Because it is 60 degrees out of phase, and one full wave cycle is 360 degrees...... you draw it 0.5 secs behind because 0.5 secs is equal to 60/360 of one full cycle. 60/360 is the same as 10/60 which is the same as 1/6. So you draw the wave 1/6th of a cycle back.
Hence, you draw it 1/6th of a cycle (0.5s) to the right of T1, because if the wave's profile is 0.5s behind T1 i.e. it's 0.5s later.
Can you attach the graph as well please?
can i have attachment?
I'm not good with computer so i will explain clearly instead of attatchment.
Your graph, which has T1 drawn on it, is showing a sinusoidal curve. A sinusoidal curve is a curve which has the same pattern throughout its duration. You can see that the curve rises up and then down. One cycle of this curve is from one up to another up.
In mathematics, one cycle(up to up) is equal to 360 degrees. just accept that, you dont have to understand. So if one cycle is 360 degrees and it is also 3 seconds, as you can see from the graph, then 0.5 seconds is also 60 degrees because 0.5 seconds is 1/6th of 3 seconds, and 1/6th of 360 degrees is 60 degrees.
Do you understand now?
Your graph, which has T1 drawn on it, is showing a sinusoidal curve. A sinusoidal curve is a curve which has the same pattern throughout its duration. You can see that the curve rises up and then down. One cycle of this curve is from one up to another up.
In mathematics, one cycle(up to up) is equal to 360 degrees. just accept that, you dont have to understand. So if one cycle is 360 degrees and it is also 3 seconds, as you can see from the graph, then 0.5 seconds is also 60 degrees because 0.5 seconds is 1/6th of 3 seconds, and 1/6th of 360 degrees is 60 degrees.
Do you understand now?
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