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.
(c) Hence find the value of V after 5 minutes,correct to 2 SFThe 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.
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?
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.theamplitude of the oscillation;
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.(ii)the mean value of V;
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.(iii) theperiod of the oscillation, in seconds;
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 radians per second.(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 ; [/quote]
M is the average value, A is the amplitude and omega is the frequency in radians per second.
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.)(ii) calculatethe value of alpha , in radians correct to 3 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.(c) Hence find the value of V after 5 minutes,correct to 2 SF