Damped Simple Harmonic Motion problem

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SUMMARY

The discussion focuses on solving a Damped Simple Harmonic Motion problem related to a 2100 kg automobile's suspension system. The spring constant (k) can be determined using Hooke's Law, where the force exerted by the spring at the new equilibrium position equals the gravitational force acting on the chassis. The damping constant (b) can be calculated using the formula b = (-2m * ln(0.65)) / t, where t corresponds to the period of oscillation, which can be derived once k is known.

PREREQUISITES
  • Understanding of Hooke's Law
  • Knowledge of Damped Simple Harmonic Motion
  • Familiarity with logarithmic functions
  • Ability to calculate oscillation periods
NEXT STEPS
  • Calculate the spring constant k using Hooke's Law for a 525 kg load.
  • Determine the oscillation period using the formula T = 2π√(m/k).
  • Use the calculated values to find the damping constant b.
  • Explore the effects of damping on oscillation amplitude in mechanical systems.
USEFUL FOR

Mechanical engineers, physics students, and professionals involved in automotive suspension design and analysis will benefit from this discussion.

davegillmour
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I'm having trouble with this problem.The suspension system of a 2100 kg automobile "sags" 7.2 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 35% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 525 kg.
 
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a) I know that k=mw^2 but don't know where I can get w from.

b)Xe^(-bt/2m)= .65X which breaks down to b=(-2m*ln(.65))/t but I don't know how/if I can't get a t.

Is there something I'm not seeing or am I totally off in my approach?

Any help is appriciated
 
davegillmour said:
a) I know that k=mw^2 but don't know where I can get w from.
You can't get omega. The way to do it is to use Hooke's law. You know the distance by which the springs get compressed when the chassis is laid down. The force exerted by the spring at the new equilibrium position must cancel the force exerted by gravity. That will give you k.
b)Xe^(-bt/2m)= .65X which breaks down to b=(-2m*ln(.65))/t but I don't know how/if I can't get a t.

Is there something I'm not seeing or am I totally off in my approach?

Any help is appriciated

The time wil correspond to one period (since it`s after one cycle). Knowing k and m you can find the period.
 

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