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
Messages
9
Reaction score
1
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
 
Physics news on Phys.org
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!
 
  • Like
Likes PeroK
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top