# Tension in the string of a simple pendulum

## 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.

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tnich
Homework Helper

## 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.

BearY
haruspex
Homework Helper
Gold Member
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?

BearY
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:
haruspex