Tension in the string of a simple pendulum

  • Thread starter BearY
  • Start date
  • #1
53
8

Homework Statement


Is the tension in the string of a pendulum, when averaged over time, larger or smaller than the weight of the pendulum? Quantify your answer. You may also assume that the angular amplitude of the oscillations is small.

Homework Equations


For tension ##T##, angular displacement of the pendulum ##\phi## and length of the pendulum ##l##
$$T = mgcos(\phi)+ ml\dot{\phi}^2$$
$$\phi = Asin(\omega t)$$

The Attempt at a Solution


I am lost at the "when averaged over time" part. I think the only way to "quantify" the answer is to examine the value of $$T_{average} - mg$$ But I don't know how to average T.
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336

Homework Statement


Is the tension in the string of a pendulum, when averaged over time, larger or smaller than the weight of the pendulum? Quantify your answer. You may also assume that the angular amplitude of the oscillations is small.

Homework Equations


For tension ##T##, angular displacement of the pendulum ##\phi## and length of the pendulum ##l##
$$T = mgcos(\phi)+ ml\dot{\phi}^2$$
$$\phi = Asin(\omega t)$$

The Attempt at a Solution


I am lost at the "when averaged over time" part. I think the only way to "quantify" the answer is to examine the value of $$T_{average} - mg$$ But I don't know how to average T.
The first thing you would need to do is express T as a function of time.
If it is not clear how to average T over time, it might help to plot T as a function of t.
Also, think about how you might use the assumption that A is small.
 
  • Like
Likes BearY
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,848
6,117
I don't know how to average T
To average over time you need to integrate wrt time. But doing that with ##T = mgcos(\phi)+ ml\dot{\phi}^2## will not be fruitful.
In approximating it for small oscillations as SHM you have already discarded third order terms; I suggest you are also expected to discard second order terms. That would approximate cos(ϕ) as ... what?
 
  • Like
Likes BearY
  • #4
53
8
To average over time you need to integrate wrt time. But doing that with ##T = mgcos(\phi)+ ml\dot{\phi}^2## will not be fruitful.
In approximating it for small oscillations as SHM you have already discarded third order terms; I suggest you are also expected to discard second order terms. That would approximate cos(ϕ) as ... what?
Sorry For the late reply I just got home.
Discarding 2nd order term, from my understanding, meaning discarding ##\frac{\phi^2}{2!}##. That would make ##cos(\phi)=1##. But isn't the variation in ##cos(\phi)## the reason this question is set up this way? Because now it's just averaging the centripetal force.

Edit: I meant it's not like I know how to solve it without discarding it but still.
 
Last edited:
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,848
6,117
Sorry For the late reply I just got home.
Discarding 2nd order term, from my understanding, meaning discarding ##\frac{\phi^2}{2!}##. That would make ##cos(\phi)=1##. But isn't the variation in ##cos(\phi)## the reason this question is set up this way? Because now it's just averaging the centripetal force.

Edit: I meant it's not like I know how to solve it without discarding it but still.
I understand your doubts, so instead write out the energy equation, without any approximations, and combine that with your tension equation. You can make a small angle approximation later.
 

Related Threads on Tension in the string of a simple pendulum

  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
2K
Replies
1
Views
10K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
6
Views
5K
Replies
3
Views
20K
Top