How Do You Calculate the Damping Constant in Damped Harmonic Motion?

[Vuldoraq]

Homework Statement

Hi all,

A hard boiled egg, with a mass m=51g, moves on the end of a spring, with force constant k=26N/m. It's initial displacement is 0.300m. A damping force F$$^{}x$$=-bv$$^{}x$$ acts on the egg and the amplitude of the motion decreases to 0.106m in a time of 5.45s.

Calculate the magnitude of the damping constant b.

Homework Equations

(1) $$\omega$$'=$$\sqrt{k/m-(b/2m)^2}$$ where $$\omega$$' is the damped angular frequency

(2) x(t)=Ae^(-bt/2m)cos($$\omega$$'t) where x(t)=displacement, A=amplitude, t=time, e=natural exponential, w'=as above.

The Attempt at a Solution

I cannot see how you can calculate the damping constant from the data given. Equation (1) above has two unknowns w' and b, while equation (2) has three x(t), b and w'.

I tried to rearrange equation (2) to get w' as the subject, but found I was unable to separate the variables. I also attempted setting w' as equal to $$\sqrt{k/m}$$, but this gave the wrong answer too. I'm really just stabbing in the dark, so some help to understand how to go about it is what I need.

I must be missing something obvious and fundamental, for instance is there another way you can calculate $$\omega$$' from the given data?

Please help!

 

[anathema]

Hello,

Thank you for your post. You are correct that there are two unknowns in equation (1) and three in equation (2), making it impossible to directly calculate the damping constant b. However, there are a few other equations that we can use to solve for b.

First, we can use the fact that the amplitude of the motion decreases to 0.106m in a time of 5.45s. This means that after 5.45s, the displacement x(t) will be 0.106m. Plugging this into equation (2), we get:

0.106 = Ae^(-b*5.45/2m)cos(\omega'*5.45)

Next, we can use the fact that the initial displacement is 0.300m. Plugging this into equation (2), we get:

0.300 = Ae^(-b*0/2m)cos(\omega'*0)

Simplifying these equations, we get:

0.106 = Acos(\omega'*5.45)

0.300 = A

Now, we can solve for A in the second equation and plug it into the first equation, giving us:

0.106 = (0.300)cos(\omega'*5.45)

Solving for \omega', we get:

\omega' = 0.106/(0.300*cos(5.45)) = 0.209 rad/s

Finally, we can plug this value for \omega' into equation (1) and solve for b:

0.209 = \sqrt{k/m-(b/2m)^2}

b = 4.97 Ns/m

I hope this helps clarify how to calculate the damping constant b from the given data. Let me know if you have any further questions.

 

[Moetasim]

Hello,

You are correct that there are two unknowns in equation (1) and three in equation (2). However, we can use both equations to solve for the damping constant b.

We can start by rearranging equation (2) to solve for A:

A = x(t)e^(bt/2m)cos(w't)

Then, we can substitute this into equation (1):

w' = sqrt(k/m - (b/2m)^2) = sqrt(k/m - (b/2m)^2) * (x(t)e^(bt/2m)cos(w't))

Since we know the values for k, m, x(t), and t, we can solve for b:

b = 2m * sqrt(k/m - w'^2) / t

Now, we can use the values given in the problem to solve for b:

b = 2(0.051kg) * sqrt(26N/m - (0.106m/5.45s)^2) / 5.45s = 0.085 Ns/m

I hope this helps! Let me know if you have any other questions.