Two parallel springs, frequency change?

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SUMMARY

When a mass M is attached to two parallel springs, the frequency of the harmonic oscillator increases by a factor of √2 compared to the frequency of a single spring. This occurs because the effective spring constant doubles when two identical springs are used in parallel. Therefore, the new frequency f' can be expressed as f' = √2 * f, where f is the original frequency of the single spring system.

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  • Knowledge of spring constants and Hooke's Law
  • Familiarity with frequency calculations in oscillatory systems
  • Basic physics concepts related to mass and force
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StephenSanders
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Ok, you guys! Please note that this is NOT a homework question but merely something that's been bothering me for a while.

We have a vertical spring with a hanging mass M. We pull the mass down and let it go, thus creating a harmonic oscillator with the frequency f.

What happens with the frequency when we attach our weight to two parellel springs of the same type as described above?

I've been thinking about this one for a while now so any help is more than welcome.

Many thanks,
Stephen
 
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Nvm, I think I finally got it. The answer is 2^(1/2)*old frequency. :)
 
Yes, the spring constant will double so the f will increase by √2.
 

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