Undamped Harmonic Motion of a rod

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Homework Help Overview

The problem involves undamped harmonic motion of a mass placed on a rod with a stationary bushing. The rod deflects 2 cm, and the bushing undergoes vertical motion defined by a sinusoidal function. Participants are tasked with determining the motion of the mass relative to the bushing or a fixed frame.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the stationary deflection of the rod and question whether the rod's motion can be assumed to follow the bushing's motion. There is confusion regarding the forces acting on the mass and how to set up the equations of motion, particularly in relation to the bushing's acceleration and the mass's response.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some suggest using Lagrangian mechanics, while others are focused on finding expressions for the forces involved. There is recognition that initial conditions may not be necessary for solving the problem, and guidance has been offered regarding the nature of the forces acting on the mass.

Contextual Notes

Participants note the lack of explicit information about damping and the properties of the rod, which may affect the setup of the problem. There is also mention of the need to consider the relative motion of the mass and the bushing, which complicates the analysis.

  • #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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