Frequency difference in two springs with different stiffness factors

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SUMMARY

The discussion centers on calculating the frequency difference between two springs with varying stiffness constants in a science fair project. The original spring has a stiffness constant "K," while the new spring has a stiffness constant of 2.2K. The frequency of oscillation is determined using the formula F = (1/2π)√(K/M). By setting up a ratio of the frequencies, it is established that the frequency of the new spring is 1.48 times larger than that of the original spring.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the formula for frequency of oscillation
  • Basic algebra for manipulating equations
  • Knowledge of mass (M) in relation to spring systems
NEXT STEPS
  • Study the derivation of the frequency formula F = (1/2π)√(K/M)
  • Explore the effects of mass on the oscillation frequency of springs
  • Investigate practical applications of spring constants in engineering
  • Learn about the relationship between stiffness and frequency in harmonic motion
USEFUL FOR

Students in physics or engineering, educators teaching mechanics, and hobbyists working on spring-based projects will benefit from this discussion.

mhmil
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Homework Statement



Your niece is working on a unique science fair project - a clock built around a small mass vibrating on a little spring. Her prototype version is working ok, but it's running too slow. Given your experience in this course, you suggest to her that a stiffer spring should oscillate faster, so she goes out and finds a new one with stiffness constant ("K") that is a factor 2.2 times larger than the prototype spring. How many times larger is the frequency of the new version than the old one? (If you think the new frequency is 2 times the old one, enter 2.0)


Homework Equations



This is where the problem is lying for me. I'm attempting to use F = 1/2pi*square root of K/M. I'm unable to find the numbers to complete the K/M at all in this question.

The Attempt at a Solution



F= (1/2pi)=(.159)
(.159)*Square root of K/M

I cannot figure out what K and M should be and am clueless as to how I find the frequency out of this question and compare the two stiffness.
 
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You don't need the actual numbers. Set up a ratio of the frequencies. Hint: If the original spring constant is K, the new one is 2.2K.
 

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