What is the distance traveled in damped harmonic motion?

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SUMMARY

The discussion centers on calculating the distance traveled in damped harmonic motion using the "suvat" equations. A participant confirms that the distance obtained is 0.5ft², but struggles with further calculations involving the maximum tension in a spring. The conversation highlights the importance of understanding the relationship between force, displacement, and the modulus of elasticity in this context. Participants collaboratively clarify the need to find the maximum extension of the spring to determine the maximum tension.

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  • Understanding of damped harmonic motion principles
  • Familiarity with "suvat" equations for constant acceleration
  • Knowledge of spring mechanics and Hooke's Law
  • Basic calculus for finding maxima and minima of functions
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  • Study the derivation and application of "suvat" equations in motion analysis
  • Learn about Hooke's Law and its implications in spring mechanics
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Students and professionals in physics, mechanical engineering, and applied mathematics who are dealing with problems related to damped harmonic motion and spring mechanics.

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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