Energy loss of damped oscillator

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Homework Help Overview

The discussion revolves around the energy loss of a damped oscillator, specifically focusing on the gravitational potential energy of a rock attached to a rope as it swings. Participants are exploring the relationship between initial and final energy states, as well as the implications of treating the rope as elastic versus inelastic.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to relate the energy equations to the scenario, questioning the definitions of variables like k and x in the context of a rope rather than a spring. There is discussion about how to calculate energy loss over multiple swings and the significance of vertical displacement in this context.

Discussion Status

Several participants are actively engaging with the problem, offering insights into potential energy loss and the need for accurate diagrams. There is an ongoing exploration of how to represent the physical setup and calculate energy changes, with some guidance provided on drawing diagrams and using trigonometric relationships.

Contextual Notes

Participants are working under constraints of missing elasticity data for the rope and are encouraged to make assumptions based on the information given. There is also a focus on ensuring clarity in diagrams to aid understanding of the problem.

  • #31
vel said:
I think we posted at the same time, lmao. I have the sine equations in #28. If I did it that all correctly and my line of thought is right, it should be PEgrav = 2.66(-9.8)(.401) = -10.45
I cannot decipher those scribblings. Please take the trouble to type in your working (as specified by the forum rules). Use the labels I made for the points to represent line lengths, e.g. OP'.
 
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  • #32
haruspex said:
I cannot decipher those scribblings. Please take the trouble to type in your working (as specified by the forum rules). Use the labels I made for the points to represent line lengths, e.g. OP'.
PP': sin(11.9) = opposite (PP')/1.45 -> 1.45sin(11.9) = .298
QQ': 1.45sin(4.1) = .103
.298 + .103 = 4.01
 
  • #33
vel said:
PP': sin(11.9) = opposite (PP')/1.45 -> 1.45sin(11.9) = .298
QQ': 1.45sin(4.1) = .103
.298 + .103 = 4.01
PP' and QQ' are horizontal distances. You are trying to find the vertical distance between P and Q.
 
  • #34
haruspex said:
PP' and QQ' are horizontal distances. You are trying to find the vertical distance between P and Q.
OP': 1.45cos(11.9) = 1.42
OQ': 1.45cos(4.1) = 1.45
OP' - OQ': 1.42 - 1.45 = -.003
 
  • #35
vel said:
OP': 1.45cos(11.9) = 1.42
OQ': 1.45cos(4.1) = 1.45
OP' - OQ': 1.42 - 1.45 = -.003
Better, but that's a bit inaccurate because you are taking the difference of two numbers that are rather close together. Keep more digits through the calculation.
 
  • #36
haruspex said:
Better, but that's a bit inaccurate because you are taking the difference of two numbers that are rather close together. Keep more digits through the calculation.
So my final would be 2.66(-9.8)(-.0274) = .7155 ? (I've got about ten minutes left to put this in; working full days and needing sleep doesn't help me solve problems, lol)
 
  • #37
vel said:
So my final would be 2.66(-9.8)(-.0274) = .7155 ? (I've got about ten minutes left to put this in; working full days and needing sleep doesn't help me solve problems, lol)

Yes, that's the energy lost over 10.5 swings, but it is not clear from post #1 exactly what you are being asked for. My guess is that it's fraction of energy lost per swing, assuming that fraction is a constant.
 

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