Finding Spring Constant for Oscillating Mass

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SUMMARY

The discussion focuses on calculating the spring constant (k) for a 2.0 kg mass oscillating on a spring, which completes 10 oscillations in 21.6 seconds. The correct frequency (f) is determined to be 0.463 Hz, calculated as 10 oscillations divided by 21.6 seconds. The spring constant is derived using the formula k = mω², where ω is the angular frequency calculated from the frequency. The conservation of mechanical energy is also applied to find the speed at the equilibrium position, confirming the calculations are accurate aside from a misunderstanding of the frequency calculation.

PREREQUISITES
  • Understanding of harmonic motion and oscillations
  • Familiarity with the formulas for frequency and angular frequency
  • Knowledge of the conservation of mechanical energy principle
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the spring constant using Hooke's Law
  • Learn about angular frequency and its relationship to frequency in oscillatory systems
  • Explore examples of conservation of mechanical energy in oscillating systems
  • Investigate the effects of mass and spring constant on oscillation frequency
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Students in physics, mechanical engineers, and anyone interested in understanding the dynamics of oscillating systems and spring mechanics.

Saladsamurai
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A 2.0kg mass attached to a spring makes 10 oscilations in 21.6s; its amplitude is .0320m

Find spring constant k. My question about this part is
frequency=(#of oscillations)/(#of seconds to complete) right?

So f=10/21.6s =2.16Hz?

So if [tex]k=m\omega^2[/tex]
[tex]k=2(2\pi*2.16)^2[/tex]
IT's the wording that gets me...

And to find speed at x=o I use conservation of mech energy

[tex]\frac{kx_m^2}{2}=\frac{mv^2}{2}+\frac{kx^2}{2}[/tex]

and I can eliminate .5kx^2 since x=0...
 
Last edited:
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10/21.6 is not 2.16 (that's 21.6/10).

Other than that, your work looks right.
 
learningphysics said:
10/21.6 is not 2.16 (that's 21.6/10).

Other than that, your work looks right.

Wow...I'm a tool...

Thanks!
 

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