Simple Harmonic Motion of spring

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SUMMARY

The discussion focuses on calculating the spring constant (k) for a mass-spring system undergoing simple harmonic motion. Given a mass (m) of 0.10 kg and a frequency (f) of 2.5 Hz, the correct spring constant is determined to be 25 N/m. The relationship between frequency and spring constant is established using the formula f = 1/(2π)√(k/m), leading to the conclusion that k can be derived from the period (T) of the motion, which is 0.4 seconds.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with the formula f = 1/(2π)√(k/m)
  • Basic knowledge of mass-spring systems
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of the formula for frequency in simple harmonic motion
  • Learn about the relationship between mass, spring constant, and oscillation period
  • Explore practical applications of Hooke's Law in mechanical systems
  • Investigate the effects of damping on simple harmonic motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to explain the principles of simple harmonic motion in mass-spring systems.

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Homework Statement


Mass is attached to a spring and set into a vibratory motion.

m=0.10 kg
f=2.5 Hz


Homework Equations



f=1/2pie{sqrt(k/m)}


The Attempt at a Solution



Trying to find k (force constant)

I tried using the equation above but I can't get the right answer.

answer: 25 N/m
 
Physics news on Phys.org
If [itex]f = 2.5 Hz[/itex] then [itex]T = 0.4 s[/itex] but we also know that [itex]T = 2\pi \sqrt{\frac{m}{k}}[/itex] so using that we can solve for k :)
 

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