Time question for v(t) = A sin(ω + Φ) V

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v(t) = 40 sin(400∏t + 150°)V

I need to find a "positive value of t for which the tension is minimal".
The answer is 1,67 ms but I don't know how to get there ?

I did "t zero" + the period of 5 ms : (-Φ/∏) + (2∏/400∏) but don't end up on it. The 5 ms addition was to jump the 1st cycle, to be at zero amplitude and be in the positive side of the X axis. I tried with half a cycle but don't get to 1,67 ms.

What am I missing ?
Thank you so much
 
Well, you can take the derivative and find when it's zero, but it's easier to do the following:
You know that sin(w) is at a minimum when w = 3*pi / 2, so simply set w = 400*pi*t + 150*pi/180 = 3*pi/2. Solve that for t.
 
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Hi,

It sounds like you should have added 2.5ms if i understand you correctly. Sometimes you have to 'jump' by half a cycle not a whole cycle.
 
Hi,

It sounds like you should have added 2.5ms if i understand you correctly. Sometimes you have to 'jump' by half a cycle not a whole cycle.

Yes, that what I meant when I said I tried with half a cycle; I added 2.5 ms but ended somewhere like 1.48 ms instead of 1.67
Sorry if it's not the exact number, my calc and book are not on the table at the moment, but yes it was close but still with a significant marging and I never have rounding errors. What to get to 1.67 ... and I'm sure it's a simple thing.
 
Hi,

Well as dougy suggested, if you take the derivative with respect to time of your function (angles in radians) you would get:
16000*pi*cos(400*pi*t+5*pi/6)

then you just set that equal to zero:
16000*pi*cos(400*pi*t+5*pi/6)=0

and now solve for t in that equation. The result is what you are after and matches your assumed result. If you ever get a negative result then just start adding half cycles until you get the value you need (the cosine wave goes through zero every half cycle).
 
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