SHM: Half Energy at Displacement

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SUMMARY

The discussion focuses on the calculations related to Simple Harmonic Motion (SHM) and energy distribution at specific displacements. When the displacement is one-half the amplitude (xm), the kinetic energy constitutes one-third of the total energy, while potential energy accounts for two-thirds. The displacement at which kinetic and potential energy are equal is found to be xm/sqrt(2). The user encountered a formatting issue with the answer submission on WebAssign, which required the correct representation of the displacement.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with energy equations in SHM, specifically U and K
  • Knowledge of amplitude and its significance in SHM
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of energy equations in Simple Harmonic Motion
  • Learn about the significance of amplitude in energy distribution
  • Explore common formatting requirements for online homework platforms like WebAssign
  • Investigate the implications of energy conservation in oscillatory systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for insights into common student challenges with energy calculations in SHM.

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Homework Statement



(a) When the displacement in SHM is one-half the amplitude xm, what fraction of the total energy is kinetic energy?

(b) What fraction is potential energy?

(c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

Homework Equations



[tex]U=\frac{1}{2}kx^2[/tex]

[tex]K+U=E[/tex]

[tex]E=\frac{1}{2}kx_m^2[/tex]

The Attempt at a Solution



I got parts (a) and (b) quite neatly, and I'm convinced that I got (c) also, only WebAssign isn't so convinced.

WORK FOR PART C:
[tex]U=\frac{1}{2}E[/tex]

[tex]\frac{1}{2}kx^2=\frac{1}{2}(\frac{1}{2}kx_m^2)[/tex]

some halfs and k's cancel, leaving

[tex]x^2=\frac{1}{2}x_m^2[/tex]

[tex]x=\frac{x_m}{\sqrt{2}}[/tex]

:eek:
 
Last edited:
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That's correct, and so is -xm/sqrt(2), since the question asked for displacement.

How did you enter the answer? Maybe it was expecting the answer in some other format.
 
neutrino said:
How did you enter the answer? Maybe it was expecting the answer in some other format.

Thanks; that was it. I entered xm/sqrt2, but they wanted xm/sqrt(2). Looks the same when you click the eyeball to ask what it'll look like!
 

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