The phase constant depends on if you use the sine or cosine function (so the phase constant in your relative equations is not the same Φ).firezap said:x(t) = Acos(wt + Φ)
x(t) = Asin(wt + Φ)
If you use cosine, then yes, it would be π/3, but should it be positive or negative?firezap said:Φ = π/3, 5.24?
firezap said:10 = 20cosΦ
0.5 = cosΦ
Φ = π/3, 5.24?
v(t)=-Aωsin(ωt+Φ).ehild said:Use the expression with pi when you give the phase.
You can decide which phase angle to use from the derivative of x(t). If x(t)=A cos (ωt+Φ) what is the sign of the derivative at t=0? What is the sigh of sin(Φ)? Which quadrant does it mean for Φ?
How do you know it's positive? It's clearly negative because of the negative sign in front of amplitude. sinΦ must be negative.ehild said:The sign of the derivative of x(t) is positive at t=0 in the plot! V(t) is positive, so what is the sign of of sinΦ??
Infinite. But there is always a Φ with |Φ|≤π/2 so I would suggest using that one.firezap said:sinΦ= 0.5. Φ=pi/6, 5pi/6
How many answers are there?
Ehild did not say that sinΦ was positive, she said that the derivative of x(t) is positive at x(0) (because the function is sloping upwards).firezap said:How do you know it's positive? It's clearly negative because of the negative sign in front of amplitude. sinΦ must be negative.
This is much easier to explain than the way I perviously thought about it!ehild said:The sign of the derivative of x(t) is positive at t=0 in the plot! V(t) is positive, so what is the sign of of sinΦ??
Look at the graph. X(t) increases at t=0, so its derivative is positive.firezap said:How do you know it's positive? It's clearly negative because of the negative sign in front of amplitude. sinΦ must be negative.