Why Should Spring Constants Be Added in a Dual-Spring System?

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SUMMARY

The discussion centers on the calculation of the effective spring constant in a dual-spring system, specifically addressing the relationship between the period of an object in Simple Harmonic Motion and the mass attached to the springs. The measured spring constants were 1.81 N/m for each spring, while calculations based on the period yielded a combined spring constant of 4.01 N/m. The conclusion drawn is that in a dual-spring configuration, the effective spring constant is the sum of the individual spring constants, resulting in a doubling effect when both springs are stretched equally.

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  • Understanding of Simple Harmonic Motion (SHM)
  • Knowledge of spring constants and Hooke's Law
  • Familiarity with graphing and interpreting linear relationships
  • Basic principles of force and mass in physics
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  • Study the derivation of the formula for the period of a mass-spring system in SHM
  • Explore the implications of combining springs in series versus parallel configurations
  • Learn about the experimental methods for measuring spring constants accurately
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of spring systems and their applications in mechanical engineering and design.

quebecois22
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Hello,

Basically, we were asked to verify the dependence of the period of an object attached on both ends by a spring upon the mass of this object in Simple Harmonic Motion. Therefore, using different masses, we calculated the period each time and made a graph where the period is a function of the root of the mass, where (2*pi)/sqrt(k) should be the slope of the line, and we had to verify the value of k obtained from that slope, to the one measured before, where we would attach the springs in a vertical manner, and hang objects and measuring the spring constant for both springs.

Now, we measured both spring constants using the latter method to be 1.81 N/m, but using all our values and the slope, we get a k value of 4.01 N/m. Now our guess was that since we had two springs, the latter is probably the sum of both springs... but why?? Are we making the right assumption?

Why, in a two spring system, should one add both spring constants when doing calculations involving k??

Thank you.. :)
 
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Because if you stretch two springs, each with the same spring constant, the same amount, you get twice the force you would have got with only one spring.
As spring constant is force per unit amount of extension, then the combined value of k will be double the single value.
 

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