What is the distance traveled in damped harmonic motion?

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Homework Help Overview

The discussion revolves around a problem related to damped harmonic motion, specifically focusing on calculating the distance traveled and understanding the forces involved in the system. Participants are exploring the application of equations related to constant acceleration and the dynamics of a spring system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use "suvat" equations to determine distance but expresses difficulty in progressing further. Some participants discuss the relationship between force, extension, and displacement, while others question the correctness of certain terms used in the equations.

Discussion Status

Participants are actively engaging with the problem, offering insights and clarifications. There is a suggestion to find the maximum tension in the spring based on the maximum extension, indicating a productive direction in the discussion. However, there is no explicit consensus on the methods being proposed.

Contextual Notes

There are indications of confusion regarding the terms used in the equations, particularly concerning the modulus of elasticity and the natural length of the spring. The original poster and others are navigating these assumptions as they seek to clarify the problem setup.

nokia8650
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http://img13.imageshack.us/img13/9091/53337497.th.jpg

Can someone please help me with the problem above? I am unable to start it. Clearly, using the constant acceleration "suvat" equations, 0.5ft^2 is the distance obtainined, however I am unable to proceed.

Thanks in advance.
 
Last edited by a moderator:
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nokia8650 said:
http://img13.imageshack.us/img13/9091/53337497.th.jpg

Can someone please help me with the problem above? I am unable to start it. Clearly, using the constant acceleration "suvat" equations, 0.5ft^2 is the distance obtainined, however I am unable to proceed.

Thanks in advance.
So you are able to do part (a), then. At any moment, t, the force on the particle is man2 times the extension of the spring. That is y and, using (a), y=(1/2)ft^2-x.

m\frac{d^2x}{dt^2}= man^2((1/2)ft^2- x)

Are you sure about "modulus of elasticity man2"? That "a" doesn't seem to fit.
 
Last edited by a moderator:
For the tension, you forgot to divide by the natural length = a, and hence the a seems to fit! Thanks a lot for the help though, it makes sense now! How would I do the last part of finding the maximum tension in the spring?

Thanks
 
Last edited:
Could anyone please help with the last part of the question please?

Thanks
 
The tension is maximum when the string's extension is at a maximum. You have an expression for the displacement. How do you find the maximum of this function?
 
Would I say y = 0.5ft^2 - (particular solution of the equation), and then find when dy/dt = 0, to find the time at which y is a max, and then subv this time back in, to find ymax, and then use the tension equation?

Thanks
 
Sounds like a good idea.
 

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