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July 5th, 2011, 08:00 AM   #1
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A double beam oscilloscope

Hi all

A double beam oscilloscope displays 2 sinusoidal waveforms A and B . The time/cm switch is on 100 s/cm and volt/cm switch on 2V/cm. The width of each complete cycle is 5 cm for both the waveforms. The height of waveforms A and B are 2 cm and 2.5 cm respectively.
Determine the phase difference ?

please help me , How I can solve like this Q ....
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July 5th, 2011, 08:25 AM   #2
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Re: A double beam oscilloscpe

Hello there r-soy,

To determine phase difference (or phase shift) between two sine waves A and B we will need some more info than what is posted.
Width and height for each signal only gives us amplitude and frequency. Is there some picture for that question which actually shows oscilloscope's output?

It could look like this one:

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July 6th, 2011, 02:14 AM   #3
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Re: A double beam oscilloscpe

My teacher told me the answer would be 36 degrees

Unfortunately, I can't get that same answer !!!!
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July 6th, 2011, 08:05 AM   #4
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Re: A double beam oscilloscpe

I think i might have found solution.
Both sine waves have same amplitude so maximum for both is 2.5 cm.
We have two sine waves A and B. They have same period / frequency and in given moment t height of sine wave A is 2.5 cm (or it's maximum) and height of sine wave B is 2 cm(not maximum yet).

Let's represent values of sine waves in given moment t with following eqations(U_max=2,5cm):
sine wave A: u(t)=U_max * sinA
sine wave B: u(t)=U_max * sinB

We know that in moment t sine wave A reaches its maximum therefore u(t)=U_max*sinA => 2,5=2,5*sinA=> sinA=2,5/2,5=1 => A = 90 degrees
For sine wave B: u(t) = 2cm, U_max= 2,5cm: 2cm = 2,5cm * sin B => sin B = 2/2,5 = 0,8 => B = 54 degrees (approximately)

Now that we know angles A and B in one moment t and it is stated that waves have same frequency we can find phase difference(P):
P=angle A - angle B = 90degrees - 54degrees = 36 degrees

Now the important thing here was that you wrote: "The height of waveforms A and B in moment t are 2 cm and 2.5 cm".I was confused at first because i thought those were the maximums for sine waves but now i understand that those were heights in certain moment.
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July 6th, 2011, 09:26 AM   #5
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Re: A double beam oscilloscpe

We could also use (assuming the same amplitude and frequency):



Divide through by A:



For simplicity, we can let:





Now we have:







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July 6th, 2011, 10:11 AM   #6
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Re: A double beam oscilloscpe

Nice! I also got 36,87 degrees because angle B is actually 53.13 deg and then P=90-53.13=36.87 degrees, but I rounded it to match r-soy's solution. Good job
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July 6th, 2011, 10:21 AM   #7
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Re: A double beam oscilloscpe

Well, once you figured out that both waves have the same amplitude A, and that A = 2.5 cm, the only real difference between what you did and what I did boiled down to using:



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July 7th, 2011, 02:39 AM   #8
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Re: A double beam oscilloscope

Yes the same amplitude was key. Maybe this problem should be more precise so we didn't have to assume this stuff?
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July 7th, 2011, 07:44 AM   #9
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Re: A double beam oscilloscope

Thank you .. I am happy now because I understand the solution ....

MarkFL , How are you ? You are absent from the scene?
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July 7th, 2011, 07:50 AM   #10
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Re: A double beam oscilloscope

I'm fine, how are you?

The part of physics I remember least about is electricity...so I haven't been able to answer many of your questions lately.
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