- #1
longball
- 5
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Could someone please look over this question
d^2i/dt^2 + 25i = 0
where i(0) =15 and di/dt(0) = 0
sketch the graph of i against t(t>0) over 2 cycles
attempt:
General Solution
i=Acos(5t) + Bsin(5t)
when i(0) = 15 is applied
A = 15
P.S
ip= Acos(5t) + Bcos(5t)
ip'= -5Asin(5t) + 5Bcos(5t)
apply di/dt (0) = 0
therefore 0=5B
so B=0
giving
i-15cos (5t)
where t = 1/5cos t = 1
Am I correct in saying that this calculation has been done correctly and the graph which has to be sketched is a cosine graph peaking and troughing at 15 and -15 with a period of 1
d^2i/dt^2 + 25i = 0
where i(0) =15 and di/dt(0) = 0
sketch the graph of i against t(t>0) over 2 cycles
attempt:
General Solution
i=Acos(5t) + Bsin(5t)
when i(0) = 15 is applied
A = 15
P.S
ip= Acos(5t) + Bcos(5t)
ip'= -5Asin(5t) + 5Bcos(5t)
apply di/dt (0) = 0
therefore 0=5B
so B=0
giving
i-15cos (5t)
where t = 1/5cos t = 1
Am I correct in saying that this calculation has been done correctly and the graph which has to be sketched is a cosine graph peaking and troughing at 15 and -15 with a period of 1