# 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