How Do You Calculate the Damping Coefficient of a Pendulum?

  • Thread starter Thread starter Punchlinegirl
  • Start date Start date
  • Tags Tags
    Damping
Click For Summary
To calculate the damping coefficient of a pendulum, the relationship between the period and amplitude decay must be utilized. Given that a pendulum of 86.0 cm has its amplitude reduced to half after 107 oscillations, the period can be derived from the string length. The differential equation governing the pendulum's motion incorporates damping proportional to the bob's speed. By solving the equations derived from the amplitude decay and the period, the damping coefficient α in Hz can be determined. This approach effectively combines the principles of harmonic motion and damping effects.
Punchlinegirl
Messages
221
Reaction score
0
A 86.0 cm pendulum is released from a small angle. After 107 oscillations the amplitude is one half of it's original value. The damping is proportional to the speed of the pendulum bob. Find the value of the damping coefficient \alpha, in Hz.

I think the period is .009 s, using 1/107, but I'm not sure if that is even right. I don't really have any idea what to do. Can someone please help?
 
Physics news on Phys.org
You can find the original period of the pendulum since you are given the string length. From there, the period has decreased by half after 107 oscillations, and there's a simple relationship between the number of oscillations and the time that's passed..
 
Punchlinegirl said:
A 86.0 cm pendulum is released from a small angle. After 107 oscillations the amplitude is one half of it's original value. The damping is proportional to the speed of the pendulum bob. Find the value of the damping coefficient \alpha, in Hz.

I think the period is .009 s, using 1/107, but I'm not sure if that is even right. I don't really have any idea what to do. Can someone please help?
Tricky question. Here's is how I would approach it:

The small angle pendulum differential equation:

\ddot\theta + \alpha\dot\theta + \frac{g}{L}\theta = 0

has solution:

\theta = \theta_0e^{-\alpha t/2}sin(\omega t) where

(1)\omega = 2\pi/T = \sqrt{g^2/L^2 - \alpha^2/4}

And you are told that:

(2)\theta = \theta_0e^{-\alpha t/2} = .5\theta_0 where t = 107T

(1) and (2) give you two equations for T in terms of \alpha so you should be able to find both.

AM
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
9
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
807
Replies
17
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
9
Views
2K
Replies
3
Views
6K
  • · Replies 5 ·
Replies
5
Views
12K