What is the value of the dampening constant?

In summary, the shock absorber of a car has elastic constant k. When the car is empty, the design is such that the absorber is critically damped. At time t=0 the absorber is compressed by an amount A from its equilibrium position and released. If after 1 second the absorber compression is reduced to A/2, what is the value of the damping coefficient?
  • #1
HiggsBrozon
7
0

Homework Statement



The shock absorber of a car has elastic constant k. When the car is empty, the design is such that the absorber is critically damped. At time t=0 the absorber is compressed by an amount A from its equilibrium position and released.
a) If after 1 second the absorber compression is reduced to A/2, what is the value of the damping coefficient? (Note you will have to solve numerically an implicit equation).


Homework Equations



x(t) = (A+Bt)e-βt where β = ω0


The Attempt at a Solution



A(t) = Ae-βt
1/2A = Ae-βt where t = 1s
ln(1/2) = -β

I know this can't be right because I'm suppose to arrive at an implicit equation. I can't seem to figure out where I'm going wrong at. Any help would be appreciated.
 
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  • #2
HiggsBrozon said:

Homework Statement



The shock absorber of a car has elastic constant k. When the car is empty, the design is such that the absorber is critically damped. At time t=0 the absorber is compressed by an amount A from its equilibrium position and released.
a) If after 1 second the absorber compression is reduced to A/2, what is the value of the damping coefficient? (Note you will have to solve numerically an implicit equation).

Homework Equations



x(t) = (A+Bt)e-βt .

The Attempt at a Solution



A(t) = Ae-βt ?
1/2A = Ae-βt where t = 1s
ln(1/2) = -β

I know this can't be right because I'm suppose to arrive at an implicit equation. I can't seem to figure out where I'm going wrong at. Any help would be appreciated.

The general equation for the displacement is

x(t) = (A+Bt)e-βt .

At t=0, x=A. What is the value of B if the velocity is 0 at t=0 (that is, the derivative of x(t) has to be zero at t=0)? ehild
 
  • #3
Thanks for you reply!

x' = (-Aβ - Btβ + β)e-βt
The velocity is 0 at t=0 so,

0 = (-Aβ + β)(1)
β=0 and A = 1

Would I then go on to use t = 1 sec at A0 = A0/2
where A0 = A+βt, the initial amplitude, or is A0 equal to the value I just solved for, A = 1?
 
  • #4
β is the damping coefficient. It is a parameter of the problem. It does not depend on the initial condition.

You have to fit the constant B to the initial condition V=0. There is a mistake in your derivation. Check the derivative of x(t) = (A+Bt)e-βt.

ehild
 
  • #5
After correcting my derivative I got,
x' = (-Aβ - Btβ + B)e-βt
Then using my initial conditions v = 0 at t = 0,

0 = (-Aβ + B)(1)
→ B = Aβ

so, x(t) = (A+Bt)e-βt where B = Aβ
and then I can use A = A/2 when t = 1s correct?
 
  • #6
Yes, it will be correct.

ehild
 

1. What is the dampening constant?

The dampening constant, also known as damping coefficient or damping ratio, is a parameter that describes the rate at which an oscillating system will lose its energy over time. It is represented by the symbol "ζ" and is used to measure the amount of resistance in a system that is undergoing damped oscillations.

2. How is the dampening constant calculated?

The dampening constant can be calculated using the equation ζ = c / (2√mK), where c is the damping coefficient, m is the mass of the system, and K is the spring constant. It can also be calculated by observing the decay of amplitude in an oscillating system over time.

3. What is the role of the dampening constant in a system?

The dampening constant plays a crucial role in determining the behavior of an oscillating system. It affects the amplitude, frequency, and stability of the oscillations. A higher dampening constant means the system will lose its energy quickly, resulting in smaller and slower oscillations. On the other hand, a low dampening constant allows the system to oscillate for a longer time with larger amplitudes.

4. How does the value of the dampening constant affect the system's response?

The value of the dampening constant can greatly impact the response of a system. If the value is high, the system will have a heavily damped response, which means it will return to its equilibrium position quickly without any oscillations. A low dampening constant will result in an underdamped response, where the system will continue to oscillate for a longer time. A critical dampening constant will provide the system with a critically damped response, which is the fastest way to return to equilibrium without any oscillations.

5. Can the dampening constant be changed in a system?

Yes, the dampening constant can be changed in a system by adjusting the parameters that affect it, such as the damping coefficient, mass, and spring constant. In some cases, external factors such as friction or air resistance can also affect the value of the dampening constant. By altering these parameters, the dampening constant can be increased or decreased, resulting in different responses from the system.

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