1. ## Graphs help?

The graph above shows the variation of voltage, V volts, with time, tseconds, for an electronic device which monitors heat transfer. It is knownthat the way V depends upon t is a sinusoidal oscillation about a non-zero meanvalue.
(a) From the graph work out
(i) theamplitude of the oscillation;
(ii) themean value of V;
(iii) theperiod of the oscillation, in seconds;
(iv) thefrequency of the oscillation, in cycles per second and in radians per second.
(b) If V(t) = M + Asin(omega[symbol wont show] t + alpha)
(i) writedown the values of M, A and omega ;
(ii) calculatethe value of alpha , in radians correct to 3 SF.
(c) Hence find the value of V after 5 minutes,correct to 2 SF

2. ## Re: Graphs help?

Hey geordi78.

Show us what you have tried: I'll give a hint for (a) (i) though:

The amplitude is the highest magnitude of a signal away from its center.

3. ## Re: Graphs help?

Originally Posted by chiro
Hey geordi78.

Show us what you have tried: I'll give a hint for (a) (i) though:

The amplitude is the highest magnitude of a signal away from its center.
V = 10
T = 2
f = .5
A = 12.5
M = 10
W = 180 deg = Pie
Alpha = 62.87 and V = -2.08

4. ## Re: Graphs help?

Not quite but I'm glad you posted an attempt.

The amplitude of a signal that is periodic is half of the total range of the signal which is 0.5*(12.5-7.5) = 2.5

T = 2, f = 0.5, M = 10, alpha = 30 degrees, A = 2.5, omega = 2 radians or 2*180/pi degrees = 360/pi degrees = 114.5916 degrees.

Do you understand how I got these answers?

5. ## Re: Graphs help?

I understand a) i,ii and iii

I dont understand the rest :S

6. ## Re: Graphs help?

theamplitude of the oscillation;
From the graph it looks like the graph goes from 7.5 to 12.5 a difference of 12.5- 7.5= 5 so the "amplitude", the distance it goes up and down from it center is half of that, 2.5.

(ii)the mean value of V;
Assuming the graph is symmetric (hard to tell since it "is not to scale"), the mean (average) value is (12.5+ 7.5)/2= 10.

(iii) theperiod of the oscillation, in seconds;
It looks like one "period" goes from a little bit less than 1 to a little bit larger than 3 so the period would be a little more than 2.

(iv) thefrequency of the oscillation, in cycles per second and in radians per second.
If the period is about "2 seconds per cycle", then the frequency is 1 cyle per 2 seconds or 1/2 cycle per second. There are [tex]2\pi[/itex] radians per cycle so the frequency in radians per second is $(1/2)(2\pi)= \pi$ radians per second.

(b) If V(t) = M + Asin(omega[symbol wont show] t + alpha)

(i) writedown the values of M, A and omega ; [/quote]
M is the average value, A is the amplitude and omega is the frequency in radians per second.

(ii) calculatethe value of alpha , in radians correct to 3 SF.
As I said before, it looks like one period starts at a little less than 1 so that would be alpha (Given the graph, especially since it is "not to scale"f, I don't see how you can get "3 significant figures.)

(c) Hence find the value of V after 5 minutes,correct to 2 SF
I don't understand the "hence"- just read the graph when t= 5. It looks like it might be about 11 but from your graph I see no way to get 2 significan figures.

7. ## Re: Graphs help?

I have this same question for an assignment. I dont actually understand questions b and c either.
I dont think the last question is worded correctly.