Period of a mass spring system with 2 spring of same K(vert)

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SUMMARY

The discussion focuses on the period of a mass-spring system when two identical springs with the same spring constant (K) are attached in parallel. The initial period of the system is given by T1 = 2π√(m/k). When a second spring is added in parallel, the effective spring constant doubles, resulting in a new period T2 = T1/√2. This confirms that the addition of springs in parallel increases stiffness, leading to a shorter oscillation period.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Knowledge of spring constants and their effects on oscillation
  • Familiarity with the formula for the period of a mass-spring system
  • Basic algebra for manipulating equations
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  • Study the effects of varying mass on the period of oscillation in mass-spring systems
  • Explore the concept of effective spring constants in parallel and series spring configurations
  • Learn about the relationship between natural frequency and oscillation period
  • Investigate real-world applications of mass-spring systems in engineering and physics
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Students and professionals in physics, mechanical engineering, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

Simon George
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A mass attached to a spring is oscillating in Simple Harmonic Motion. If an other spring of same sprinc constant is attached parrallel to the other spring, what is the period of this new system (as a function of the initial period).

Here's what I did and have no idea if this is right:

For the first period, T1= 2*pi*sqrt(m/k)

For the second period, T2= 2*pi*sqrt(m/(2k))
T2=(2*pi*sqrt(m/k))/sqrt(2)
T2=T1/sqrt(2)

Is that right?
 
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Sanity check:
Two springs in parallel are stiffer than either spring separately.
The stiffer the spring(s), the higher the natural frequency.
The higher the natural frequency, the shorter the period.

You can use a similar line of reasoning for changes in the mass.

And your math looks correct.
 

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