Graph of sinusoid

1. Feb 2, 2016

amsi

Okay I feel really stupid for getting wrong omega for such a trivial problem.

1. The problem statement, all variables and given/known data

Find the function of e(t) = A* sin(wt + fi)

2. Relevant equations

3. The attempt at a solution
A = 0.9 V
at t = 0 e(t) = -0.35, and therefore:
-0.35 = 0.9 * sin (w*0 + fi) / * arcsin
arcsin(-7/18)= fi
fi = -22.89 degrees = -0.4 rad

at t = 26 us , A = 0 V, and therefore:
0 = 0.9 * sin (w*26u -0.4 ) / *arcsin
0 = w*26 u - 0.4
w = 0.4/26 * 10 ^- 6
w = 15384 rad / s which is not correct

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2. Feb 2, 2016

Staff: Mentor

Your phase difference between the points is not 0.4 rad. Be careful with multiples of pi. The -0.4 rad would mean you are approaching the (arbitrary) phase 0 soon. That is not true. Zero is where the curve goes from negative to positive, that is more than pi rad to the left or a 0.4 rad less than pi to the right.

3. Feb 2, 2016

amsi

Yes this is also bothering me - 0.4 rad = - 22 degrees ,it would mean sinusoide is 'moved' to the right for 22 degrees out of phase, also I look it this way like judging from picture, sine is moved from the phase of function -Asin(wt) = A*sin(wt+180) to the left for 22 degrees so it's Asin(wt+202 degrees) ,then I get that omega is w= 136197 rad/s

4. Feb 2, 2016

Staff: Mentor

Your phase advances from "0.4 after a zero crossing" to "pi after this zero crossing", so pi-0.4.
Alternatively, from pi+0.4 (an alternative solution for the sine) to 2 pi (also an alternative solution), which gives the same difference.

5. Feb 4, 2016

amsi

I get that w= 105445.87 s^-1 ,but I don't understand why did I get as a result 0.4 rad,how can we prove this mathematically by equations,not just looking from graph.

6. Feb 4, 2016

Staff: Mentor

Well, a sine reaches the same value (apart from 1 and -1) twice per oscillation. You have to "look at the graph" (or know the derivative at this point) to figure out which one applies to your case.