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Waves Find the phase constant from the graph.

  • #1
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Homework Statement


http://i.imgur.com/TZcJkjH.jpg
Find the phase constant

Homework Equations


x(t) = Acos(wt + Φ)
x(t) = Asin(wt + Φ)
w = 2pi/T

The Attempt at a Solution


I see A = 20cm. T = 4s. f = 1/4 = 0.25Hz. w = 2pi/4 = 1.57Hz.
10 = 20cosΦ
0.5 = cosΦ
Φ = π/3, 5.24?
 

Answers and Replies

  • #2
Nathanael
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x(t) = Acos(wt + Φ)
x(t) = Asin(wt + Φ)
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).

Φ = π/3, 5.24?
If you use cosine, then yes, it would be π/3, but should it be positive or negative?
 
  • #3
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sinΦ= 0.5. Φ=pi/6, 5pi/6
How many answers are there?
I'm guessing -π/3 since you asked.
 
  • #4
ehild
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10 = 20cosΦ
0.5 = cosΦ
Φ = π/3, 5.24?
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?
 
  • #5
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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?
v(t)=-Aωsin(ωt+Φ).
sign of derivative is negative. sign of sinΦ is positive. That means in quadrant 1 and 2.
 
  • #6
ehild
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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??
 
  • #7
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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??
How do you know it's positive? It's clearly negative because of the negative sign in front of amplitude. sinΦ must be negative.
 
Last edited:
  • #8
Nathanael
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sinΦ= 0.5. Φ=pi/6, 5pi/6
How many answers are there?
Infinite. But there is always a Φ with |Φ|≤π/2 so I would suggest using that one.

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 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).
 
  • #9
Nathanael
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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??
This is much easier to explain than the way I perviously thought about it!

I suppose sin(ωt+Φ) would be even easier, because you don't even need check the derivative (just check if x(0) is positive or negative)

Thanks for this perspective, I never thought about it in this way.
 
  • #10
ehild
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How do you know it's positive? It's clearly negative because of the negative sign in front of amplitude. sinΦ must be negative.
Look at the graph. X(t) increases at t=0, so its derivative is positive.
 
  • #11
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the answer is -π/3 for cos and π/6 for sin.
 

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