Time taken for energy to drop 95% in damped SHO

In summary: In any event, I'm glad to see the original post is correct. Good work!In summary, the conversation discusses a post addressing a physics problem involving a damped oscillating spring. The post provides helpful information for solving part B of the problem, which involves calculating the frequency and percentage decrease of the oscillation's amplitude. The conversation also discusses different methods for solving part C, which involves finding the time interval for the system's energy to decrease to 5% of its initial value. The final answer of 21.2 seconds, which is double the solution in the back, is deemed correct. The only discrepancy is with the units given for the damping coefficient, which should be Ns/m rather than Nm/s.
  • #1
Phynos
31
4
There is a post about the same problem here: https://www.physicsforums.com/threads/damped-oscilating-spring.12838/

It was helpful for solving part B.

1. Homework Statement


A 10.6kg object oscillates at the end of a vertical spring that has a spring constant of 2.05x10^4 N/m. The effect of air resistance is represented by the damping coefficient b = 3.00Ns/m.

(a) Calculate the frequency of the damped oscillation.

(b) By what percentage does the amplitude of the oscillation decrease in each cycle?

(c) Find the time interval that elapses while the energy of the system drops to 5.00% of it's initial value.

Homework Equations



1. x = Ae^(bt/2m) cos(wt + phi)

2. (From post mentioned above): %Difference = 1 - e^(bt/2m) * 100%

The Attempt at a Solution



I have completed part (A): frequency is 7Hz and (B): Percent drop is 2%.

For part (C) I tried solving the second equation above for t, then subbing in difference of 0.95.

I end up with t = -(2m/b) ln(1-D)

solving when D = 0.95 gives me:

t = 21.2s

This is exactly double the solution in the back. Why is it not the same? I didn't insert a factor of two anywhere, the two in the equation is meant to be there.

Is the back wrong? Or am I making a stupid mistake somewhere? It's more likely that I've made a mistake but I've been looking at it awhile now and I'm stumped.

---

I also tried multiplying the time taken to drop 2% (The period, sqrt(k/m) ) by 47.5 but then I end up with:

t = 6.97s

I realized after this would not work because the decrease in energy is not linear, hence the exponential term in the equation.
 
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  • #2
Hi Phynos, Welcome to Physics Forums.

Looking at your method I can't find fault. t = 21.2s seems like a correct value to me. Of course, I stand ready to be corrected by someone more clever than I :smile:

I do have one nit to pick, and that's with the units of the given damping coefficient. I think they should be Ns/m rather than Nm/s. Usually the air resistance force is modeled as being proportional to velocity in this type of problem, so b*V should yield Newtons.
 
  • #3
Thanks.

[Ns/m][m/s] = N

The units are in Ns/m in my post. Perhaps you misread, unless you meant they should not be? But I think they are correct.
 
  • #4
Phynos said:
Thanks.

[Ns/m][m/s] = N

The units are in Ns/m in my post. Perhaps you misread, unless you meant they should not be? But I think they are correct.
Ah. My apologies. I think I must have been confused by Klion's post (which you linked to).
 

1. What is a damped SHO?

A damped SHO, or damped simple harmonic oscillator, is a system that exhibits periodic motion that decreases in amplitude over time due to an external force or resistance.

2. How is the time taken for energy to drop 95% calculated?

The time taken for energy to drop 95% in a damped SHO can be calculated using the equation t = ln(1/0.05) * (2π/ω), where t is time, ln is the natural logarithm, and ω is the angular frequency of the oscillator.

3. What factors affect the time taken for energy to drop 95% in a damped SHO?

The time taken for energy to drop 95% in a damped SHO is affected by the damping coefficient (b), the mass of the oscillator (m), and the spring constant (k) of the system. A higher damping coefficient or mass will result in a shorter time, while a higher spring constant will result in a longer time.

4. Can the time taken for energy to drop 95% be altered?

Yes, the time taken for energy to drop 95% in a damped SHO can be altered by changing the damping coefficient, mass, or spring constant of the system. Additionally, adjusting the initial amplitude and phase of the oscillator can also affect the time taken.

5. Why is it important to understand the time taken for energy to drop 95% in a damped SHO?

Understanding the time taken for energy to drop 95% in a damped SHO is important in various fields of science and engineering, such as in the design and analysis of mechanical systems, electrical circuits, and acoustic oscillations. It also helps in predicting the behavior and stability of these systems over time.

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