# Motion with damping constant

1. Oct 8, 2009

### flame_m13

1. The problem statement, all variables and given/known data
I'm a little ashamed to post this, because I bet I'm missing something obvious, so try not to laugh...

A freely rolling freight car weighing 10^4 kg arrives at the end of it's track with a speed of 2 m/s. At the end of he track is a snubber consisting of a firmly anchored spring with k = 1.6 x 10^4 kg/s^2. The car compresses the spring. If friction is proportional to the velocity, find the damping constant b$$_{c}$$ for critical damping. Find the maximum distance by which the spring is compressed (for b = b$$_{c}$$). Show that if b > b$$_{c}$$ the car will come to a stop, but if b < b$$_{c}$$ the car will rebound and roll back on the track. (the car is not fastened to the spring. As long as it pushes on the spring, it moves according to the harmonic oscillator equation, but it will not pull on the spring).

2. Relevant equations
$$\frac{d^2 x}{dt^2} + 2\beta\frac{dx}{dt} + \omega^{2}x = 0$$

Also, for critical damping, $$\omega$$$$^{2}$$ = $$\beta$$$$^{2}$$

$$\beta$$ = b/2m, where b is the damping constant.

3. The attempt at a solution
from the above, i calculated the critical damping constant b = 2.53 x 10^4 (1/s). I'm not sure how to find how far the spring is compressed due to that, and I don't know how I could attempt the remaining parts of the problem. Hints?

Last edited: Oct 8, 2009
2. Oct 8, 2009

### chrisk

Solve the differential equation with the given initial conditions. An elementary differential equations text will show how to solve the equation and address the underdamped, critically damped, and over damped conditions.

3. Oct 8, 2009

### flame_m13

well, i know the general solution for the differential equation is
$$x(t)= e^{-t * \beta}[A_{1}exp(\sqrt{\beta^2 - \omega^2}) + A_{2}exp(-\sqrt{\beta^2 - \omega^2})$$

Should I use this to find the compression of the spring?

4. Oct 8, 2009

### chrisk

It's a little easier to express the solution for x in terms of cosine and sine. Your expression is for the distance the spring compresses so use it to find the compression distance. Hint: what is the velocity of the object when the fully compressed?

5. Oct 8, 2009

### flame_m13

the velocity of the freight car would be zero when the spring is fully compressed.

so if the velocity is zero when the spring is completely compressed, then the first equation would be
$$\frac{d^2 x}{dt^2} + 2\beta(0)+ \omega^{2}x = 0$$
= $$\frac{d^2 x}{dt^2} + \omega^{2}x = 0$$

which is the equation for an undamped simple harmonic oscillator. the general solution of this is

$$x(t) = A cos(\omega t)$$
so the compression would be x = A cos (1.6 t) ??? Am I going in the right direction? Or am I just completely off track?

I've attempted the second part of the question, about the damping constant changing. for b < b$$_{c}$$, $$\sqrt{\beta ^2 - \omega ^2}$$ is imaginary, so
x(t) = Ae$$^{- \beta t}$$cos ($$\omega$$t-d). So for this case, since the car isn't attached to the spring, it would bounce back and then keep going back on the track. Is my logic/ math correct? Thanks again.

6. Oct 8, 2009

### chrisk

Eliminating the velocity term of the differential equation gives a different equation of motion, that of a simple oscillator. Use the solution you found, express it in terms of cosine and sine terms rather than complex exponents (makes the math more clear). Determine the constants for the cosine and sine terms from the initial conditions when the object first contacts the spring. Differentiate x(t) with respect to time to obtain the v(t) (with friction proportional to velocity). Set v(t) = 0 to find the time of travel then place t in x(t) to find the distance traveled.