# Homework Help: Why is a pendulum only simple harmonic motion for small angles?

1. Aug 11, 2014

### 21joanna12

1. The problem statement, all variables and given/known data
Hi all!
I'm slightly confused about pendulums and simple harmonic motion. In my text book, it says that a pendulum only exhibits simple harmonic motion when the angle is small (<10 degrees). I was wondering why this is, using equations if possible.
Without the math, I think this is the case because the part of the pendulum's motion that exhibits SHO is its horizontal displacement from the equilibrium position (i'm not sure about this though. Does the up/down motion also exhibit SHO or can the whole motion, not just the horizontal motion, be described as SHO) and in that case the restoring force can be approximated by the tangential force, which is -mgsinθ. And then only for small angles is sinθ≈θ, and so only then is the acceleration proportional to the displacement as in SHO... I really think this is wrong.

2. Relevant equations

3. The attempt at a solution
I'm trying to find the acceleration of the pendulum bob at any point, and so far I have that the resultant force perpendicular to the velocity of the bob is 2mgcosθ-2mgcos(θmax) and the tangential force is mgsinθ, but i'm not even sure where I am going with this...

I don't really get pedulums, so any help on this would be much appreciated!!!

2. Aug 11, 2014

### Orodruin

Staff Emeritus
You have the tangential force. In order to have harmonic motion, the force should be proportional to the displacement, in your case the angle. Instead, your force is proportional to the sine of the angle.

The reason you have harmonic oscillations for small angles is that the sine of a small angle is essentially equal to the angle.

3. Aug 11, 2014

### 21joanna12

Thanks Orodruin for your reply. I believe I mentioned the sine approximatuon in my post, however there is something else that is troubling me- textbooks, and your earlier post, seem to treat this as the only approximation. But we made an approximation from the beginning in assuming thatthe restoring force is ONLY the tangential component of the weight, when in fact there would be a component of the string tension in it as well (which would of course decrease with a decreasing angle) and radial component of gravity. What if the approximations cancel out?

Also, is the only part of a pendulum that models SHO the horizontal displacement? What about treating the pendulum as a whole and considering its position vector from its equilibrium position, not just horizontal displacement?

Sorry for more questions....

4. Aug 11, 2014

### Matterwave

Make sure you draw the force diagram.

By definition a tension can not act in a direction perpendicular to the string, and that direction is exactly the tangential direction.

If you wanted to work in x-y coordinates then you will have contributions of the tension in both x and y directions, as well as contributions of the gravity in both x and y directions. The trick with a pendulum is to realize that because the bob must remain at the end of the string (I assume the string is taunt! or, we can assume a rigid rod instead of a string), then we can reduce the number of equations we must solve by moving to r,θ coordinates. In this case the tension acts entirely in the r-direction, a direction we don't care for since r is constant.

5. Aug 11, 2014

### PaulDirac

Exact solution for calculating the period of a pendulum makes use of expanding the sin function appears in the differential equation of motion in terms of the first three mostly effective expressions, which bear most of the weight of sin function. But as soon as a low approximation is concerned, one can replace sin which appears in the question with x, which precludes a massive expression for the equation of a simple pendulum.

6. Aug 11, 2014

### 21joanna12

I think I may not have been very clear when I was describing what I meant... what I was trying to get at was that as the bob approaches the equilibrium position, then the string is not perpendiicular to the horizontal (I didn't mean the tangential direction) and therefore will be a part of the horizontal restoring force (I think).

Thank you for your help! ^_^

EDIT: sorry Matterwave! I realise that I started talking about a different question II posted about tension in strings, so I have removed the second part of my previous comment!

7. Aug 11, 2014

### SteamKing

Staff Emeritus
This article derives the equation of motion for the simple pendulum. It discusses the simplification made for small angles in the sine term, which allows for a solution which exhibits SHM. For the more general case where the angles are not small, elliptic functions must be introduced to provide a solution, which diverges from SHM.

http://en.wikipedia.org/wiki/Pendulum_(mathematics)

8. Aug 11, 2014

### Orodruin

Staff Emeritus
Yes, the string tension is changing and will always be constraining in its own direction. This means providing exactly the force necessary to cancel out any radial component.

9. Aug 11, 2014

### 21joanna12

Orodruin, I'm really sorry for bothering you again! But in my textbook, it says that 'as the bob passes through equuilibrium, the tension T acts directly upwards. Therefore the resultant force on the bob at this instant is

T-mg= mv2/l

where v is the speed as it passes through equilibrium'

This seems to me to be suggesting that the tension does not cancel out the radial component of the weight...

10. Aug 11, 2014

### Orodruin

Staff Emeritus
This is true, my wording was a bit hasty and not precise. The tension is going to be exactly as large as it needs to be too keep the pendulum in circular motion, i.e., providing the difference between the gravitational component in the radial direction and the centripetal force necessary (mv^2/r).

11. Aug 11, 2014

### Matterwave

It's much easier to not worry too much about the tension, because it gives you only the force in the radial direction. Which we don't care about in this problem, as I mentioned earlier. We just know it works out.

It is the force in the tangential, theta, direction that we care about.

At points where the bob is moving, it is accelerating in the radial direction. This is the same basic principle as in circular motion, except the velocity is now time dependent. So the tension must provide for both the weight of the bob, as well as provide the centripetal force necessary to keep the bob moving in a circle.

12. Aug 12, 2014

### 21joanna12

I think I get it now... the radial force is irrelevant when considering the motion?