- #1
Ethan Godden
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Homework Statement
A particle with a mass(m) of 0.500kg is attached to a horizontal spring with a force constant(k) of 50.0N/m. At the moment t=0, the particle has its maximum speed of 20m/s and its moving to the left. Find the minimum time interval required for the particle to move from position x=0 to x=1.00m.
Homework Equations
General Equation of Motion: x(t)=(A)cos(ωt +Φ) where A is the amplitude or maximum position, ω is the angular frequency, and Φ is the phase constant
ω=√(k/m)
vmax=ωA
The Attempt at a Solution
ω=√(k/m)=√(50/0.5) = 10.0 rad/s
vmax=ωA → A=(vmax/ω)=20/10=2.00m
Since at t=0, the particle is at its maximum speed, the phase constant must make cosine=0 as the particles maximum speed is at position x=0, so the phase constant must be π/2.
Equation of Motion: 2cos(10t+π/2)=x(t)
I believe the solution from here it to make x(t)=1.00m as the particle is at x=0 at t=0, so the time at x(t)=1.00m is the time it takes to go from x=0 to x=1.00.
2cos(10t+π/2)=1.00 → cos-1(1/2)=10t+π/2 → 10t= π/3 - π/2 → t= -π/60=-0.0524s.
I believe the answer is just the absolute value of this. When I input t=0.0524s, i get x(t)=-1.00. Shouldn't this be equivalent to the time it takes to get to +1.00m.
Apparently, the answer is 0.105s which is noticed is double the magnitude of my answer. I am pretty sure this is not a coincidence. It would be greatly appreciated if someone could tell me where I am going wrong.
Thank You,