You almost had the correct answer in post #1. The trick I mentioned in post #17 is probably your simplest solution. B/A=2.5/10.33. It should be a simple matter to find ## \phi ## and write out the answer. And in the form ## y=C\cos(\omega t-\phi) ## the shift is to the right for positive ## \phi ##Enochfoul said:Thanks for all of your help and for the other guys help too. This question has been doing my head in.
Any chance you could have a look at the impedance question I posted. I can't decide if the final answer should be positive or negative.NascentOxygen said:It's right.
Either way, the answer should be the same.
I have the same question myself and have followed it up until this point, but i am confused as to wear the value 125 and -516.5 come from?Enochfoul said:Ok I've had another bash.
So If C = 2.5 sin 50t + 10.33 cos 50t
the dC/dt = 125 cos 50t - 516.5 sin 50t (diff of sin 50t is 50 cos 50t and diff of cos 50t is -50 sin 50t)
At a maximum dC/dt is 0
0 = 125 cos 50t - 516.5 sin 50t
rearrange and divide by cos 50t this give tan 50t = 2.5/10.33 giving 50t as 13.6 deg.
So 13.6 deg converted to radians would be 0.237.
0.237 ÷ 50 = 0.00474 which would be the time the amplitude of 10.6 would occur?
Using you tip of changing the graph to "Maximum of" it comes out as http://www4f.wolframalpha.com/Calculate/MSP/MSP72722a107agdc0he8ga0000405a140d63h2172g?MSPStoreType=image/gif&s=42&w=496.&h=33.
Im guessing this is correct then?
The function of time was differentiated. The values are a result of that operation.joe forrester said:I have the same question myself and have followed it up until this point, but i am confused as to wear the value 125 and -516.5 come from?
gneill said:The function of time was differentiated. The values are a result of that operation.
Differentiating trig functions sin and cos:
##\frac{d}{dt} Asin(ω t) = Aωcos(ω t)##
##\frac{d}{dt} Bcos(ω t) = -Bωsin(ω t)##