Damped Oscillator homework

In summary: There are two ways to get the correct answer: 1) use the value of ##\gamma## that they give you, but then you cannot eliminate the linear term as they do in their solution; 2) use the value of ##\gamma## I get, which means that their solution is incorrect.
  • #1
jbrussell93
413
38

Homework Statement


A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on which to mount the platform so that the platform will settle to a new equilibrium position 0.2 m below its original position as quickly as possible after the impact without overshooting.

a)Find the spring constant k and the damping constant b of the dashpot.
b)Find the time required for the platform to settle within 1 mm of its final position.



Homework Equations


$$mg=kx$$
$$m\ddot{x}+b\dot{x}+kx=mg$$
$$\omega=\sqrt{\frac{k}{m}}$$
$$\gamma=\frac{b}{2m}$$

Answers:

a) ##k=49000kgm/s^{2}##, ##b=70700 kgm/s##

b)##0.076sec##


The Attempt at a Solution



I am having trouble finding the value of ##b##. I assumed that since the question asks for the quickest amount of time without overshooting, this would be a critical damping problem with ##\omega=\gamma## but when I solve there is overshooting. The only way that I am able to get the answer for part b) is to take away from forcing function ##mg##.
 
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  • #2
Additionally, when ##\gamma>\omega## my solution is of the form
$$x(t)=e^{-\gamma t}(c_{1}e^{\sqrt{\gamma^{2}-\omega^{2}}t}+c_{2}e^{-\sqrt{\gamma^{2}-\omega^{2}}t})+\frac{mg}{k}$$

Instead of the usual form for critical damping when ##\omega=\gamma##
$$x(t)=e^{-\gamma t}(c_{1}+c_{2}t)+\frac{mg}{k}$$
 
  • #3
I don't see how they got their value of ##b## either.

Given the initial conditions and parameters, it doesn't look like there's a way to not have the platform overshoot the new equilibrium position. For a critically damped oscillator, you have
$$x(t) = [x_0 + (v_0+\gamma x_0)t] e^{-\gamma t}.$$ With the assumption that the upward direction is positive, to have no overshoot, the coefficient of the linear term has to be positive. If not, ##t## will eventually get large enough so that x(t) becomes negative, which means that the mass has overshot the new equilibrium point.
 
  • #4
vela said:
Given the initial conditions and parameters, it doesn't look like there's a way to not have the platform overshoot the new equilibrium position. For a critically damped oscillator, you have
$$x(t) = [x_0 + (v_0+\gamma x_0)t] e^{-\gamma t}.$$

That is what I ended up with for the situation with ##F(t)=0## but wouldn't ##F(t)=mg##? Assuming downward positive and ##x=0## at the equilibrium position, I took

##x_0=-.2 m##
##v_0=14 m/s##
##\gamma=70 s^{-1}##

which causes ##(v_0+\gamma x_0)t## to equal 0. There would be no overshooting in this case, just a smooth decay to zero. By doing this, I got the correct answer for part b) but I still believe that there should be a forcing function somewhere in there.
 
  • #5
F(t)=mg only serves to shift the equilibrium point downward.

How'd you get γ=70 1/s? I get that it's equal to 7 1/s.
 
  • #6
vela said:
F(t)=mg only serves to shift the equilibrium point downward.

Hmm this explains why I can't figure out how to get the mass to settle to x=0 with the forcing function included... I asked my professor about this one and he told me that there should be a forcing function but I don't see how because all other terms go to zero leaving only that constant. It should be zero.

vela said:
How'd you get γ=70 1/s? I get that it's equal to 7 1/s.

By solving $$(v_0+\gamma x_0)t=0$$ but that is partly why I'm so confused. I was under the impression that in order to use that solution x(t), ##\omega## had to equal ##\gamma## in which case ##\gamma=7## and the mass overshoots.
 
  • #7
jbrussell93 said:
Hmm this explains why I can't figure out how to get the mass to settle to x=0 with the forcing function included... I asked my professor about this one and he told me that there should be a forcing function but I don't see how because all other terms go to zero leaving only that constant. It should be zero.
The solution with the forcing function is ##x(t) = mg/k + (c_1+c_2 t)e^{-\gamma t}##. As ##t \to \infty##, ##x(t) \to mg/k##. What you can do is define ##x' = x - mg/k##, and x'(t) will satisfy the homogeneous equation. The solution I wrote above is really for x'(t), not x(t).

With your sign convention, if you use x(t), the initial conditions are x(0)=0 and ##\dot{x}(0) = 14\text{ m/s}##, and the equilibrium position is at x=0.2 m. If you use x'(t), the initial conditions are x'(0)=-0.2 m and ##\dot{x}'(0) = 14\text{ m/s}##, and equilibrium is at x'=0.

By solving $$(v_0+\gamma x_0)t=0$$ but that is partly why I'm so confused. I was under the impression that in order to use that solution x(t), ##\omega## had to equal ##\gamma## in which case ##\gamma=7## and the mass overshoots.
You're right. The numbers given in the problem don't work out, so I wouldn't worry about it if I were you. Whoever wrote the problem made an error, perhaps using the wrong value of ##\gamma## when solving it.
 

1. What is a damped oscillator?

A damped oscillator is a physical system that exhibits oscillatory motion, but with a gradual decrease in amplitude over time. This decrease in amplitude is due to the presence of a damping force, which dissipates energy from the system.

2. How is the behavior of a damped oscillator described mathematically?

The behavior of a damped oscillator can be described using a second-order differential equation known as the damped harmonic oscillator equation. It takes into account the mass of the system, the spring constant, and the damping coefficient to predict the position of the oscillator over time.

3. What is the difference between an underdamped, critically damped, and overdamped oscillator?

An underdamped oscillator experiences a gradual decrease in amplitude and an oscillatory motion, while a critically damped oscillator approaches equilibrium quickly without oscillating. An overdamped oscillator experiences a slow approach to equilibrium without oscillating.

4. How do you solve a damped oscillator equation?

To solve a damped oscillator equation, one can use a variety of methods such as the Laplace transform, the method of undetermined coefficients, or Euler's method. The specific method used will depend on the complexity of the equation and the initial conditions given.

5. What are some real-life examples of damped oscillators?

Some real-life examples of damped oscillators include a swinging pendulum, a car's suspension system, and a guitar string. In each of these cases, the motion of the system is affected by a damping force, leading to a decrease in amplitude over time.

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