# Simple Pendulum undergoing harmonic oscillation

• Safder Aree
In summary, the time average of the tension in the string of the pendulum is smaller than mg by a factor of cos(theta).
Safder Aree

## Homework Statement

Is the time average of the tension in the string of the pendulum larger or smaller than
mg? By how much?

## Homework Equations

$$F = -mgsin\theta$$
$$T = mgcos\theta$$

## The Attempt at a Solution

I'm mostly confused by what it means by time average. However from my understanding of the question the answer should be yes it is smaller than mg. Since $$T=mgcos\theta$$.

Therefore, it is smaller by a factor of cos(theta) unless the angle is 0 or pi in that case it is equal to mg. Does that make sense?

I would suggest you consider the fact that the mass on the end of the pendulum does not have any net vertical acceleration across the period of one cycle.

jambaugh said:
I would suggest you consider the fact that the mass on the end of the pendulum does not have any net vertical acceleration across the period of one cycle.
Sorry, I'm not sure I follow.

Picture the swinging pendulum. Picture the force due to tension acting on the bob, the mass m. That force as a vector changes direction a bit as the pendulum swings. As a vector it has a vertical component and horizontal component. Think about your question in terms of the time average of components, vertical and horizontal.

haruspex
jambaugh said:
That force as a vector changes direction
Tension is not a vector. However, I would not rule out the possibility that the question setter intended it to be taken as such.
@Safder Aree , there is something else to consider.
What are the forces on the mass when at the lowest point of the swing? What is its acceleration there?

haruspex said:
Tension is not a vector.
which is why I didn't say it was. I said the force due to tension on the bob is a vector.
The tension is the magnitude of the named vector and the direction is defined by the angle of the pendulum as it swings.

@haruspex the force on the mass when at the lowest point of the swing will be a maximum but it will not give unambiguous information about the time average.

@Safder Aree Here's another hint. Imagine replacing the pendulum with an equal mass suspended from a spring with spring constant such that the mass spring system oscillates twice as fast as the pendulum... but if you look at their vertical components only, they match. The mass spring is modeling just the vertical component of the pendulum. The forces on the mass in the mass spring system will be just (about) the same as the vertical component of the force on the mass in the pendulum system.

(the frequency of the mass spring needs to be double because the pendulum reaches max height and min height twice per cycle.)

[EDIT: Actually this spring example may not be the easiest way there... consider this fact: Since the system is periodic, the time (vector) average of all forces on the mass must be zero over a period since its motion is periodic. All forces including the force of gravity. Think about what that means when you express the total force (not always zero but averaging to zero) as a sum of gravity's constant force and the string's (or spring's in my other example) force on the mass.

jambaugh said:
which is why I didn't say it was. I said the force due to tension on the bob is a vector.
No, but you wrote this:
jambaugh said:
Think about your question in terms of the time average of components, vertical and horizontal.
Which is not going to work because tension is not a vector.

jambaugh said:
the force on the mass when at the lowest point of the swing will be a maximum but it will not give unambiguous information about the time average.
Consideration of that position should lead to the realisation that a contributor to tension has been omitted.

Safder Aree said:

## Homework Statement

Is the time average of the tension in the string of the pendulum larger or smaller than
mg? By how much?

## Homework Equations

$$F = -mgsin\theta$$
$$T = mgcos\theta$$
No, the tension is not mgcos(θ),
The bob moves along a circle, so the net radial force should be equal to the centripetal force.
Safder Aree said:

## The Attempt at a Solution

I'm mostly confused by what it means by time average.
Time average is the integral to one period divided by the length of a period. For small diplacements, the angle of the string changes according to SHM: ## θ=Asin(\frac{2π}{T}t+φ)## . If the bob starts from the equilibrium position, φ=0.
You know how the the time period T of the bob is calculated from the length of the pendulum and the mass of the bob. The velocity of the bob is v=(dθ/dt) L. Once you get the time dependence of the velocity, you can write out the time dependence of the tension, and you can integrate it for one period.

ehild said:
the angle of the string changes according to SHM:
ehild said:
you can integrate it for one period.
We are not told these are small oscillations, and the integral is quite nasty for the general case.
Fortunately we are not asked to find the average tension, merely how it compares with mg.

@jambaugh had the best idea - no integrals required - but I misunderstood the direction he was going because he mentioned averaging the horizontal component. The key is to average just the vertical component and consider how the existence of a horizontal component affects the average magnitude.

haruspex said:
We are not told these are small oscillations, and the integral is quite nasty for the general case.
Fortunately we are not asked to find the average tension, merely how it compares with mg.

@jambaugh had the best idea - no integrals required - but I misunderstood the direction he was going because he mentioned averaging the horizontal component. The key is to average just the vertical component and consider how the existence of a horizontal component affects the average magnitude.

I actually forgot to mention that we are to assume small oscillations. I'm still quite confused on where to proceed with this question. How would you go about averaging the vertical component.

Would T:
$$Tcos(\theta) = mg$$
$$Tsin(\theta) = F$$

Thus
$$T = \frac{mg}{cos(\theta)}$$

Safder Aree said:
I actually forgot to mention that we are to assume small oscillations.
Ok, but using @jambaugh's method you don't need to assume that.
If the average vertical force over a swing is Fy,2π, and the period is T, what would be the net change in vertical momentum?
This approach is the easiest way of answering how the average tension compares with mg.

If you want to take the approach of calculating what the average tension is (assuming small oscillations):
Safder Aree said:
Would T:
$$Tcos(\theta) = mg$$
No, that's not right either. Please don't make wild guesses.
You were closer with:
Safder Aree said:
$$T = mgcos\theta$$
which is true at each end of the swing.
For a general instant of the swing, as I mentioned in post #5 and @ehild in post #8, you are overlooking centripetal force.
Draw a free body diagram of the general position. What is the net force in the radial direction? What is the acceleration in that direction?

haruspex said:
We are not told these are small oscillations, and the integral is quite nasty for the general case.
Fortunately we are not asked to find the average tension, merely how it compares with mg.
The title of the thread is about "simple pendulum undergoing harmonic oscillation." And it is also asked: by how much is it smaller or greater than mg?

As for integration, my method uses only the fact that the integral of a periodic function is zero for a full period.
The problem can be solved also with conservation of energy, and then using small-angle approximation.
@ Safder Aree: But the first thing is to know how the tension depends on the position of the bob.

Last edited:
ehild said:
it is also asked: by how much is it smaller or greater than mg?
You're right - I missed that. But then the question becomes, to which order of approximation?
I feel that being told it is SHM doesn't quite address that. Taking it as SHM, we might come up with an average tension that differs from mg in the θ02 term. In that case, starting afresh, not taking it as SHM, but still applying a small angle approximation, we could get a different θ02 term.

haruspex said:
You're right - I missed that. But then the question becomes, to which order of approximation?
I feel that being told it is SHM doesn't quite address that. Taking it as SHM, we might come up with an average tension that differs from mg in the θ02 term. In that case, starting afresh, not taking it as SHM, but still applying a small angle approximation, we could get a different θ02 term.
I think a first order approximation would be quite satisfactory for the average tension.

Both methods should give the same second order term in θ0.

Last edited:
ehild said:
I think a first order approximation would be quite satisfactory for the average tension.
Even if that is just mg? Don't we need the first nonzero term beyond the zeroth order?

haruspex said:
Even if that is just mg? Don't we need the first nonzero term beyond the zeroth order?
How do you know that the first nonzero term is quadratic in θ0?

ehild said:
How do you know that the first nonzero term is quadratic in θ0?
That is the result I got.

haruspex said:
That is the result I got.
You are right, it is quadratic in θ0, but the same with both methods.

ehild said:
You are right, it is quadratic in θ0, but the same with both methods.
I get different results, but my two methods need not be the same as your two.

haruspex said:
I get different results, but my two methods need not be the same as your two.
My first method was assuming SHM : θ=θ0sin(ωt) with ##ω=\sqrt{g/L}##
The other method was with conservation of energy.
I used the approximation sin(x)≈x, cos(x)≈1-x2/2.

ehild said:
My first method was assuming SHM : θ=θ0sin(ωt) with ##ω=\sqrt{g/L}##
The other method was with conservation of energy.
I used the approximation sin(x)≈x, cos(x)≈1-x2/2.
Same here, but I can see other possible points of divergence.
No time right now, but I'll post my solutions in a private message, tomorrow probably.

haruspex said:
Same here, but I can see other possible points of divergence.
No time right now, but I'll post my solutions in a private message, tomorrow probably.
I used time average.

## 1. What is a simple pendulum undergoing harmonic oscillation?

A simple pendulum undergoing harmonic oscillation is a system where a small mass (called the bob) is suspended from a fixed point by a light string or rod. When the bob is displaced from its equilibrium position and released, it will oscillate back and forth due to the force of gravity.

## 2. What factors affect the period of a simple pendulum undergoing harmonic oscillation?

The period of a simple pendulum undergoing harmonic oscillation is affected by the length of the string, the mass of the bob, and the acceleration due to gravity. A longer string, larger mass, and higher gravitational acceleration will result in a longer period.

## 3. What is the formula for calculating the period of a simple pendulum undergoing harmonic oscillation?

The formula for calculating the period of a simple pendulum undergoing harmonic oscillation is T = 2π√(L/g), where T is the period in seconds, L is the length of the string in meters, and g is the acceleration due to gravity in m/s².

## 4. How does amplitude affect the motion of a simple pendulum undergoing harmonic oscillation?

The amplitude of a simple pendulum undergoing harmonic oscillation is the maximum displacement of the bob from its equilibrium position. A larger amplitude will result in a wider range of motion and a longer period, while a smaller amplitude will result in a narrower range of motion and a shorter period.

## 5. What is the relationship between the period and frequency of a simple pendulum undergoing harmonic oscillation?

The period and frequency of a simple pendulum undergoing harmonic oscillation are inversely related. This means that as the period increases, the frequency decreases, and vice versa. The frequency is the number of oscillations per second and is equal to 1/T, where T is the period in seconds.

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