Undamped Harmonic Motion of a rod

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The discussion centers on solving a problem involving the undamped harmonic motion of a mass on a rod with a stationary bushing. The key issue is how to account for the forces acting on the mass, particularly the relationship between the motion of the mass and the bushing's vertical motion. Participants clarify that the rod's deflection indicates its properties and that initial conditions are not critical for finding a periodic steady-state solution. The conversation also emphasizes the need to define the net force on the mass correctly, considering both gravitational and spring forces, and to express the mass's motion relative to the bushing. Ultimately, the solution involves setting up the equations of motion correctly to account for these interactions.
  • #31
ah yeah, duh!

x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k
 
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  • #32
Or acutally, couldn't I just find that with

x(0)=-mg/k= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

so

A=-mg/k
 
  • #33
Feodalherren said:
Or acutally, couldn't I just find that with

x(0)=-mg/k= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

so

A=-mg/k
No, that only works at t=0. Over time, the average of that sum of trig functions is zero, not -mg/k.
 
  • #34
So then by the principle of superpositioning it should be
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k

but then I arrive back at my original issue - how do I find A and B. I only have one initial condition?
 
  • #35
Feodalherren said:
So then by the principle of superpositioning it should be
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k

but then I arrive back at my original issue - how do I find A and B. I only have one initial condition?
No, you have two. At time zero it is at rest in the equilibrium position.
 
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