The discussion revolves around determining the phase constant from a provided graph of a wave function. The subject area includes wave mechanics, specifically the mathematical representation of wave motion using trigonometric functions.
The discussion is active, with various interpretations of the phase constant being explored. Participants are engaging in reasoning about the signs of derivatives and the implications for the phase angle. Some guidance has been offered regarding the use of derivatives to determine the correct quadrant for the phase constant.
There is an ongoing debate about the signs of the phase constant and the implications of the graph's behavior at specific points. Participants are also considering the constraints of the problem, such as the requirement to express the phase angle within certain bounds.
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.