Finding Spring Stiffness and Effective Mass for Suspended Beam

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SUMMARY

The discussion focuses on calculating the spring stiffness and effective mass for a suspended beam system. The beam, when unloaded, has a vertical vibration period of 0.83 seconds, and when loaded with an additional 50 kg, the period increases to 1.52 seconds. The effective spring stiffness was determined to be 608 N/m, using the formula for equivalent spring constant in parallel springs (keq=2k). The maximum dynamic tension during sudden stops of the suspended safe was also analyzed, emphasizing the importance of initial conditions in dynamic systems.

PREREQUISITES
  • Understanding of Newton's Second Law
  • Familiarity with harmonic motion and vibration periods
  • Knowledge of spring constants and equivalent spring stiffness
  • Basic principles of dynamic tension in suspended systems
NEXT STEPS
  • Study the derivation of the formula for equivalent spring constants in parallel systems
  • Learn about dynamic tension calculations in suspended loads
  • Explore the effects of mass loading on vibration periods
  • Investigate the relationship between frequency and spring stiffness in oscillatory systems
USEFUL FOR

Mechanical engineers, physics students, and anyone involved in the analysis of dynamic systems and vibration mechanics will benefit from this discussion.

rock.freak667
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Homework Statement


A cable is used to suspend a 400 kg safe. It is being lowered at 6 m/s, when it stops suddenly. Find the maximum dynamic tension and the frequency of vibration. k=2(105 N/m


Homework Equations



Newton's 2nd law

The Attempt at a Solution



Well I easily found the frequency using ω=k/m and got the correct answer.

If I find the resultant force in the direction of mg, I get mg-T=ma. I am not given a, I am given, v.


Homework Statement


A uniform beam is suspended by 2 springs A & B. Unloaded, the period of vertical vibration is 0.83s. When 50kg is loaded onto its centre, the period is 1.52s. Find the spring stiffness of each spring and the effective mass of the beam


Homework Equations





The Attempt at a Solution



The springs would be in parallel so

keq=2k

ω=√keq/m ⇒(2π/0.83)^2 (50) = 2k, but this does not give me the correct answer. k=608 N/m. I get 1432.66 N/m
 
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1) Right, the velocity they give you will be an initial condition that you can use with the general equation for a mass on a spring. Max tension should be at max deceleration.

2) It gives you the period for the unloaded beam. Once the beam is loaded with more mass it will have a different period. :)
 
Last edited:

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