How Many Oscillations to Reduce Amplitude by 1000 in a Damped System?

[frostking]

Homework Statement

A mass spring system has the following parameters: damping constant is 1.00 Ns/m spring constant is 1.00N/m and mass is 1.00 kg The mass is displaced from equilibrium and released. Through what minimum number of oscillations must the mass move in order to reduce the initial amplitude by a factor of 1000 or more?

Homework Equations

x(t) = A e ^(-bt/2m) angular accel W = (k/m - b^2/4m^2)^(1/2) T = 2pi/w

The Attempt at a Solution

I computed w and got 0.86 I know that e ^(-bt/2m) is equal to or less than 1/1000 But this is where I get stuck. Some number times w needs to be less than 1/1000 the original amplitude but...I cannot see how to get there. I know from the solution sheet that the answer is two but can someone please show me how to get there? Thanks, Frostking

 

[Delphi51]

frostking, I'm the retired high school teacher here to give a helping hand when I can. It looks like you've got an exponential that needs to be inverted into a log. When you have e^x = .001, you can take the natural log of both sides to get x = ln(.001). I hope that will be useful!

 

[kerimek]

I would approach this problem by first understanding the concept of damping harmonic motion. In this system, the damping constant represents the amount of energy lost per unit time due to friction or other non-conservative forces. This means that as the mass oscillates, its amplitude will decrease over time due to the dissipation of energy.

Using the given parameters, we can calculate the natural frequency of the system (w) using the equation w = sqrt(k/m - b^2/4m^2). Plugging in the values, we get w = sqrt(1/1 - 1/4) = sqrt(3/4) = 0.866.

Now, we can use the equation x(t) = A e ^(-bt/2m) to calculate the amplitude at any given time t. Since we are looking for the minimum number of oscillations required to decrease the amplitude by a factor of 1000, we can set up the following inequality:

A e ^(-bt/2m) < A/1000

Dividing both sides by A and taking the natural logarithm of both sides, we get:

-ln(1000) < -bt/2m

Solving for t, we get:

t > (-2m/b) ln(1000)

Substituting in the values for m and b, we get:

t > (-2/1) ln(1000) = -2 ln(1000) = 13.815 seconds

This means that the mass must complete at least 2 full oscillations (since it takes 2 oscillations to complete one full cycle) in order to reduce the amplitude by a factor of 1000 or more. This is because at t = 13.815 seconds, the amplitude will be approximately 0.001 times the initial amplitude, satisfying the given condition.

Therefore, the minimum number of oscillations required is 2.