Oscillations: The position is modeled as x(t)=Acos(ωt+ϕ)

AI Thread Summary
The discussion focuses on solving for the phase angle and position of an oscillating object modeled by the equation x(t)=Acos(ωt+ϕ). The object oscillates at a frequency of 2.44 Hz, and at t=0.246 s, it reaches its maximum amplitude of +0.15 m. Participants clarify that the phase angle ϕ should be calculated as negative, leading to a value of approximately 2.51 radians. The importance of verifying the calculations by substituting values back into the original equation is emphasized. The conversation concludes with a successful resolution of the phase angle and appreciation for collaborative assistance.
jdmaxwell02
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Homework Statement
A object oscillates with a frequency of 2.44 Hz. At t=0.246 s, the object is located at its amplitude A=+0.15 m.
The position is modeled as x(t)=Acos(ωt+ϕ)

Determine the phase angle (positive value between 0 and 2π):

a. Determine the phase angle (positive value between 0 and 2π):
b. Determine the position of the object at t=0 s.
Relevant Equations
w=2pi/T
What I have done so far:
Since x=A, cos(ωt+ϕ) must equal 1
cos-1(1)=0 so -ωt=ϕ
ω=2pi/(1/f)=15.3
ωt=3.7=ϕ
 
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jdmaxwell02 said:
Homework Statement:: A object oscillates with a frequency of 2.44 Hz. At t=0.246 s, the object is located at its amplitude A=+0.15 m.
The position is modeled as x(t)=Acos(ωt+ϕ)

Determine the phase angle (positive value between 0 and 2π):

a. Determine the phase angle (positive value between 0 and 2π):
b. Determine the position of the object at t=0 s.
Homework Equations:: w=2pi/T

What I have done so far:
Since x=A, cos(ωt+ϕ) must equal 1
cos-1(1)=0 so -ωt=ϕ
ω=2pi/(1/f)=15.3
ωt=3.7=ϕ

You have the equation ##\phi = -\omega t##, but you seem to have taken ##\phi## to be positive.
 
PeroK said:
You have the equation ##\phi = -\omega t##, but you seem to have taken ##\phi## to be positive.
So would ϕ be 3.77?
 
jdmaxwell02 said:
So would ϕ be 3.77?

Why not plug ##\omega, t## and ##\phi## into your equation and see whether you get ##\cos(\omega t + \phi) = 1##?
 
ϕ is 2.51 rad. Plugged it into my calculator. Thanks for the help!
 
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