Calculating Amplitude and Energy in a Damped Oscillator: A Homework Question

[UMDstudent]

Homework Statement

An oscillator with a mass of 520 g and a period of 0.500 s has an amplitude that decreases by 1.00% during each complete oscillation.

PART A : If the initial amplitude is 10.2cm , what will be the amplitude after 43.0 oscillations?

PART B: At what time will the energy be reduced to 64.0% of its initial value?

Homework Equations

Xmax(t) = Ae^(-bt/2m)

The Attempt at a Solution

I've spent the past hour reviewing the problem and reading the textbook but I cannot seem to get a grip on our to successfully complete either PART A or B. Starting with PART A, we have the unknown b and if you solve for b, you will have the unknown of xmax (amplitude after 43 oscillations). I'm frustrated with the problem and I'm hoping to get some help.

We know the amplitude, the time, and the mass. We don't know the damping constant (b)

Thanks,

UMDstudent

 

[JaWiB]

Perhaps you're making it too difficult by focusing on the exponential decay equation. If the amplitude decreases by 1% each oscillation, then after one oscillation the amplitude is .99*10.2cm; after two it is .99*(.99*10.2cm); then .99*(.99*.99*10.2cm) and so on.

Assuming the period is constant then the second part shouldn't be too difficult; you might first calculate the number of oscillations that would lead to 64% of the energy, then calculate the time.

 

[UMDstudent]

Great recommendation. The first part is correct; essentially you just take .99 multiplied by the original amplitude. As for the second part, I believe the answer is 22.2 seconds but mastering physics says I am wrong. I came to 22 seconds by : 44.4 oscillations to reach 64 % of initial value (6.5 of 10.2). Divided it by 2 (since it takes .5000 seconds). Any ideas?

 

[JaWiB]

Great recommendation. The first part is correct; essentially you just take .99 multiplied by the original amplitude. As for the second part, I believe the answer is 22.2 seconds but mastering physics says I am wrong. I came to 22 seconds by : 44.4 oscillations to reach 64 % of initial value (6.5 of 10.2). Divided it by 2 (since it takes .5000 seconds). Any ideas?

64% of initial value of amplitude is not equal to 64% of initial value of energy

 

[UMDstudent]

So we solve for total energy; in this case : E = U + K = 1/2kA^2 + 1/2mv^2. Our unknowns become both k & v so once we solve for these unknown's and add them into this equation, we would need 64% of this value?

 

[JaWiB]

Not quite. U + K is the total energy, but they both change during the motion. For a harmonic oscillator, U = 1/2k*x^2 where x is the displacement from equilibrium. A is the maximum displacement from equilibrium, but what is the kinetic energy at this displacement?

 

[UMDstudent]

The kinetic energy at the turning point (the maximum displacement from equilibrium) would be equal to zero.U = 1/2*K*A^2. Wouldn't this be the total energy of the equation?

 

[UMDstudent]

U = 1/2 (82.1)(.102)^2

k = omega^2 * Mass = 82.1

U = .427
------------------------------------------------
K = 1/2 mv^2
K = 1/2 (.52)(4.01)^2
K = 4.18

v = square root (k/m * A)
------------------------------------------------
Total Energy = K + U

TE = 4.61

64% of TE = 2.95.

TE - 64% = 1.66 ?(Sorry for double post)

 

[UMDstudent]

BUMP... I need some clarification to see if I am correct?