Underdamped Harmonic oscillator with applied force

AI Thread Summary
An underdamped harmonic oscillator is analyzed under the influence of an applied force, leading to the equation of motion incorporating mass, spring constant, and damping resistance. The solution for position x(t) is expressed as a combination of homogeneous and inhomogeneous parts, focusing on the inhomogeneous response. The amplitude A and phase shift ∅ are derived from the applied force and system parameters, but the initial conditions x0 and v0 need to be integrated into the solution. Clarification is sought on how these initial conditions affect the final expression for x(t) and whether the amplitude varies with time. The discussion emphasizes the need for a comprehensive understanding of how initial conditions influence the oscillator's behavior.
nbram87
Messages
8
Reaction score
0

Homework Statement



An underdamped harmonic oscillator with mass m, spring constant k, and damping resistance c is subject to an applied force F0cosωt.
(a) [analytical] If, at t = 0, x = x0 and v = v0, what is x(t)?


Homework Equations



Ωinitial = √(k/m)

The Attempt at a Solution



Fnet = -kx - cv + F0cos(Ωt) = ma
x(t) = xh(t) (homogeneous) + xi(t) (inhomogeneous) so we will only be left with the inhomogeneous part
x(t) = Acos(Ωt - ∅)

A = (F0/m) / ((√(Ωinitial^2 - Ω^2)^2 + 4gamma*Ω^2))

∅ = tan-1((2*gamma*Ω)/(Ωinitial^2 - Ω^2))

I have all that but I am confused where x = x0 amd v = v0 come into play and how to plug it all into a x(t) equation
 
Physics news on Phys.org
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 '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