Two parallel springs, frequency change?

AI Thread Summary
When a mass is attached to two parallel springs, the effective spring constant doubles, which affects the frequency of the harmonic oscillator. The new frequency is calculated as the square root of 2 times the original frequency. This results in an increase in frequency due to the combined stiffness of the two springs. The discussion confirms that the frequency change is a direct consequence of the altered spring constant. Understanding this relationship is crucial for analyzing systems involving multiple springs.
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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