Calculate Damping Coefficient of 50.0g Egg

[mlee]

A 50.0-g hard-boiled egg moves on the end of a spring with force constant . It is released with an amplitude 0.300 m. A damping force acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.Calculate the magnitude of the damping coefficient . Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures

pls who can help me?
thanx

 

[arildno]

How should Newton's 2.law of motion look like?

 

[mlee]

i think it is:
-kx-bv=ma

 

[arildno]

That's correct!
Now, what type of solutions have you learned that this differential equation has?

 

[Pyrrhus]

See it as

-kx - b \frac{dx}{dt} = m\frac{d^2 x}{dt^2}

 

[Pyrrhus]

You're right, thanks alridno

 

[mlee]

v= dx/dt and a= d^2/dt^2

 

[mlee]

but what is the answer of d^2/dt^2 then?

 

[arildno]

mlee:
Any progress at what sort of solutions your equation has?

 

[mlee]

uh not really...;(

 

[arildno]

Now, I'd like you try a solution of the form:
x(t)=Ae^{rt} (A and r constants)
What condition must be placed on "r" in order for this to be a solution.
Please post your work.

 

[mlee]

Asin(wt)+Bcos (wt)

 

[arildno]

Asin(wt)+Bcos (wt)

This is a solution of an UNDAMPED, harmonic oscillator.
Your oscillator is NOT undamped; try my approach, and post your work.

 

[mlee]

Ae-bt/2mCos(ω't + φ)

 

[mlee]

Ae^(bt/2m)*cos(w't+φ)

 

[arildno]

You lack a minus sign in your exponential!
Now, knowing
a) The initial displacement
and
b)That the initial velocity is zero
How can you determine A,\phi

Besides, what is your value of "w"?

 

[mlee]

ω = sqrt(k/m)
ω' = √((k/m) - (b²/4m²))

 

[arildno]

Now, so how does your initial conditions determine A,\phi?

 

[mlee]

i don't know how to find phi

 

[mlee]

and w' = 5*10^2-(b^2/1*10^-2)
is that right?

 

[arildno]

Now, initially we have:
A\cos\phi=A_{0}
where A_{0} is the initial displacement.
In order to find the position, we differentiate:
\frac{dx}{dt}=Ae^{-\frac{bt}{2m}}(-w'\sin(w't+\phi)-\frac{b}{2m}\cos(w't+\phi))
Hence, for t=0, we must have:
0=A(-w'\sin(\phi)-\frac{b}{2m}\cos(\phi))

 

[arildno]

Furthermore, in order to solve the problem, remember that:
Ae^{\frac{-bt}{2m}} is the AMPLITUDE as a function of time..

 

[mlee]

cos phi is 0.333?
is that right?

 

[arildno]

You get the equations:
A\cos\phi=A_{0}
and
tan\phi=-\frac{b}{2mw'}

 

[mlee]

b is unknown

 

[arildno]

You're right!
While I have a method to determine 'b', I don't think this is what has been intended.
I think that it has been assumed (incorrectly!) that the amplitude function is:
A_{0}e^{\frac{-bt}{2m}}
where A_{0},m are known quantities.
Hence, it is simple to determine 'b' from this.
(Just plug in the proper t-value and set your amplitude equal to the given value)
I think this has been the intention; the equation you may derive for 'b' is not easy to solve.