# Find the dampening variable

1. Sep 23, 2007

### miann

1. The problem statement, all variables and given/known data

A egg with a mass 0.0500kg is suspended from a spring with a k = 25.0N/m... initially it starts off with a displacement of x = 0.300m but a force F = -bv acts against the egg and after t= 5.00s the displacement is x = 0.100m

2. Relevant equations

F = -bv
F = -kx
-kx = ma

w = $$\sqrt{}K/m$$

ma = -kx - bv

3. The attempt at a solution

Well our prof didnt show up for class... but the homework is still due and i really have no idea were to start... i have never worked with damping... but i found out that

b = kg/s

Because F/v = kg/s

So if u calculate F at x = 0.300m you get 7.5N and then 2.5N for 0.1m...

thats a difference of 5... and that would be 1N/s loss when divided by time... which would make sense for the overall damping effect... (well i'm guessing... again never learned anything about damping) but if that is right... what 'v' would i use... because it is always changing as the spring looses energy...

So ya, I dont knoe if i'm even going in the right direction... but i am completely stumped on this one.

2. Sep 23, 2007

### Proggle

This is a differential equation. Treat it as such.

3. Sep 23, 2007

### miann

all I can think of is

x(t) = Acos(wt + $$\phi$$)

v(t) = -wAsin(wt + $$\phi$$)

a(t) = -w$$^{2}$$Acos(wt + $$\phi$$)

4. Sep 24, 2007

### miann

Well... i can set it up as

-kx - b$$dx/dt$$ = m$$d^{2}x/dt^{2}$$

5. Sep 24, 2007

### miann

but i cant think of the second equation so i can solve...

6. Sep 24, 2007

### miann

sorry i havent done differential equations for about a year.. and even then they were easy ones... i'm trying to figure it out but i have no idea were to start.

7. Sep 24, 2007

### Mindscrape

Typically you want guess exponentials. x = Ae^(rt)

8. Sep 24, 2007

### Proggle

Try
http://www.krellinst.org/UCES/archive/modules/diffeq/node4.html [Broken]

Last edited by a moderator: May 3, 2017
9. Sep 24, 2007

### miann

kx = -bv

kx - -b$$\frac{dx}{dt}$$

$$\int^{0.1}_{0.3}$$$$\frac{dx}{x}$$ = -$$\int^{t}_{0}$$$$\frac{k}{b}$$dt

ln(0.1) - ln(0.3) = $$\frac{k}{b}$$(5.0s)

1.1 = $$\frac{(25.0N/m)}{b}$$(5.0s)

b = $$\frac{(25.0N/m)(5.0s)}{1.1}$$

b = 113.64 N$$\bullet$$s/m

is that right?

Last edited: Sep 24, 2007