wayne Posted January 24, 2007 Posted January 24, 2007 Hi, Given: v=40 cos(100pi t + pi/3) Question: what is the min number of msec that the function must be shifted to the left if the expression for v is 40 cos(100pi t)? therefore I have the equation: 40 cos(100pi (t+t0) + pi/3) = 40 cos(100pi t) hence, 100pi (t+t0) + pi/3 = 100pi t However at that point I decided to check the answers making sure I was on the right track and I was not. According to my book: the equation to start was: 40 cos(100pi (t+t0) + pi/3) = 40 cos(100pi t +2pi) with the 2pi we end up with a t0 positif but the way I did it, t0 is negatif. Is that the reason why they added the 2pi ? Thanks for your answer(s)
m4rc Posted February 9, 2007 Posted February 9, 2007 Your answer would have been valid if the question did not specify that the graph must be shifted to the left. However since it does specify this, they are looking for the first value of t0 that is positive. You will find that "40 cos(100pi t +2pi)" is equivalent to "40 cos(100pi t)" because it is shifted by one full period. You can always add one (or multiple) period(s) to get a phase factor with the correct sign.
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