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Homework Help: The Power Cable Voltage Sinusoid Word Problem Question (How do you obtain C?)

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data
    The voltage V in a power cable varies sinusoidally with time t. At t = 1, the voltage is 12 volts. The voltage rises to its maximum level of 20 volts by t = 6. At t = 11, the voltage returns to 12 volts and then begins to fall. The minimum voltage is 4 volts.

    2. Relevant equations
    y = (+/-)A*cos(Bx (+/-) C)+D or y = (+/-)A * sin(Bx (+/-) C)+D

    3. The attempt at a solution
    Finding A:
    The use of (M-m) / 2 yields the amplitude, which is (20-4)/2 = 8.
    So, A = 8.

    Finding B:
    The use of 2pi / B yields the period. So, assuming that the period of the average voltage function is 10 (11-1=10), I have 2pi / B = 10.
    2pi/10 = B
    pi/5 = B
    So the period is pi/5.

    Finding C:
    Unsure of how to proceed in this step, please advise.

    Finding D:
    The use of (M+m)/2 yields the vertical shift, which is (20+4)/2 = 12.
    So, D or vertical shift in this case is 12.

    My guess at the equation of this sinusoid is y = -8cos((pi/5)*x + C) + 12.

    Any assistance is helping me understand how to obtain C would be invaluable to me.
  2. jcsd
  3. Jan 8, 2008 #2
    You figured out A and D right I believe(8 and 12)

    If you ever forgot what these things meant, you could just use some given info and the kinda "framework" wave, V=A*cos(B*t+C)+D (remember V is a function of t here, not x) so V=8*cos(B*t+C)+D, you can do whatever you want with the negatives, like if you did -8 instead of 8, that's peachy but it'll change the phase shift(C) by some amount. There's more than one right answer depending on if you use sin, cos, + or - in various places

    so at t=1 V=12, so 12=8*cos(B+C)+12
    at t=6 V=20, so 20=8*cos(B*6+C)+12

    two equations, two unknowns! I assume you found B correctly and used the correct equation, note that you mispoke and said the period is pi/5 and also B, which is untrue. In that case you just go with 12=8*cos(pi/5+C)+12 and solve for C

    More properly though, C is how much it's shifted left or right, and you can immediately eyeball the answer from the first two sentences in your question and knowing that 12 is the middle value for the wave
  4. Jan 8, 2008 #3
    I meant to say B = pi/5.

    The period is 10 in this particular case, due to it taking a total amount of 10 time units to complete a full cycle.

    I'll attempt to solve for C as you suggested, I thought that there was another equation / option.
  5. Jan 8, 2008 #4
    Err, the period should be 20...

    So, B = 2pi/20 -> B = pi/10.

    The horizontal shift C would be 1 to the right...

    The correct equation would be something like: V = 8*sin(pi/10 * (t-1)) + 12

    Can anybody validate that?
  6. Jan 9, 2008 #5
    Mostly, you got C the right way but then fiddled with the equation's set up.

    First: Remember it was all V=A*sin(B*t+C)+D but you threw the t in with the C for some reason

    Just to make a quick semi-related point, notice that you also could've used cos instead of sin, and that would've changed your answer for C, but it would've still been right! That's easy to know if you remember sin is the same as cos, but shifted......pi/2? That wouldn't make it easier, but just to show that that weird equation with all the + or - and sin or cos stuff is just saying use whatever you want

    NOTE: I don't think this is right, gimme a sec...
    ok here's a graph of what V=8*sin(pi*t/10-1)+12
    http://www.gomath.com/cgi-bin/plot.py?eq=8*sin%28pi*x%2F10-1%29%2B12&eq=&eq=&eq=&eq=&autoscale=1&xlow=&xhigh=&ylow=&yhigh= [Broken]

    Amplitude's right, vertical shift is right, but at t=1 it's clearly NOT 12 volts like it should be, and so on. See how easy these are to check? I checked and your equation for B is slightly wrong, http://en.wikipedia.org/wiki/Angular_frequency (B is angular frequency, if you wanna know the cool names. C is phase shift. Now impress your teacher!)
    Last edited by a moderator: May 3, 2017
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