Simple Harmonic Motion - Finding max speed of an oscillator

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SUMMARY

The discussion focuses on calculating the maximum speed of a 300g oscillator using principles of simple harmonic motion. The oscillator's speed is given at two displacements: 95.4 cm/s at 3.0 cm and 71.4 cm/s at 6.0 cm. By applying the conservation of energy and manipulating the equations of motion, the maximum speed was determined to be 102.2 cm/s. The spring constant (k) was calculated to be 44.48 N/m, and the total energy (E_tot) was found to be approximately 0.157 J.

PREREQUISITES
  • Understanding of simple harmonic motion equations
  • Familiarity with conservation of energy principles
  • Knowledge of spring constant (k) calculations
  • Ability to manipulate trigonometric functions in physics equations
NEXT STEPS
  • Study the derivation of the maximum speed formula in simple harmonic motion
  • Learn about energy conservation in oscillatory systems
  • Explore the relationship between displacement, velocity, and acceleration in harmonic oscillators
  • Investigate the effects of mass and spring constant on oscillation frequency
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for practical examples of simple harmonic motion calculations.

KEØM
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Homework Statement


A 300g oscillator has a speed of 95.4 cm/s when its displacement is 3.0 cm and 71.4 cm/s when it displacement is 6.0 cm. What is the oscillator's maximum speed?


Homework Equations


x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
v = -[tex]\omega[/tex]Asin([tex]\omega[/tex]t + [tex]\phi[/tex])
vmax = [tex]\omega[/tex]A
v = [tex]\omega[/tex](A^2 - x^2)^1/2

The Attempt at a Solution


I manipulated the fourth equation into

[tex]\omega[/tex]A or vmax = (v^2 + (x[tex]\omega[/tex])^2)^1/2

so if I get [tex]\omega[/tex] from somewhere else then I can use it to solve it for vmax but I can't see a way of doing that with the given info.
 
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Hmm... the fact that the mass is given gives me an idea. Try approaching the question from the conservation of energy, where Etot = 1/2 mv2 + 1/2 kx2. Equate Etot of the two cases would solve for k, and then one could find the value of Etot.

The next step is fairly simple: at maximum speed, Etot = 1/2mvmax2
 
Thank you very much for your help.

For a value of k I got 44.48 N/m

and for E[tex]_{}tot[/tex] I got about .157J

and then solving for v[tex]_{}max[/tex] I got 1.022 m/s or 102.2 cm/s which makes sense.

Thanks again for your help.

KE0M
 

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