# OCR M3 Textbook Question (Exercise 6 Question 7)

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Hi there,

I'm having trouble with the following Mechanics question:

"A pendulum swings through an angle a on either side of the vertical. The length of the thread is l, and the mass of the bob is m. Find an expression for the tension in the thread when the bob is in its lowest point. Find also the tension predicted by using the SHM approximation".

The first part is fine - I use a conservation of energy equation to get v^2, and then resolve radial forces and use the fact radial acceleration = (v^2)/r to solve for T, getting mg(3-2cos(alpha). The next part is what gives the problem.

I understand that the SHM approximation is when you get the equation d(theta)/dt^2 = -sqrt(g/l)sin(theta), and use the fact that sin(theta) is approximately theta to get an equation of SHM form. However, this equation doesn't involve T. Additionally, under the SHM model, we take the only force to be opposing the direction of motion - e.g transverse - and so would surely assume T = mg...?

The answer in the back of the book is mg(1+a^2). This can quite easily be obtained from the answer I got through Conservation of Energy (mg(3-2cos(a)), using the approximation cosa = 1 - 0.5a^2 for small a. However this isn't using SHM in any form ...

So in short - am I misunderstanding what we mean when we treat a pendulum as SHM (so we DO have a radial component...?), and thus is there some method to get this answer which uses SHM more directly?

Sorry for my poor explanation - if my workings so far are unclear I can scan them in.

EDIT: Thinking about it, I see why T = mg doesn't work. We're not saying the displacement of the bob obeys SHM, but our angle made with the vertical, which is why radial acceleration doesn't have to = 0. This leads me to believe that perhaps the point of the exercise WAS just to use the cos small angle approcimation - but this isn't SHM?!

I'm having trouble with the following Mechanics question:

"A pendulum swings through an angle a on either side of the vertical. The length of the thread is l, and the mass of the bob is m. Find an expression for the tension in the thread when the bob is in its lowest point. Find also the tension predicted by using the SHM approximation".

The first part is fine - I use a conservation of energy equation to get v^2, and then resolve radial forces and use the fact radial acceleration = (v^2)/r to solve for T, getting mg(3-2cos(alpha). The next part is what gives the problem.

I understand that the SHM approximation is when you get the equation d(theta)/dt^2 = -sqrt(g/l)sin(theta), and use the fact that sin(theta) is approximately theta to get an equation of SHM form. However, this equation doesn't involve T. Additionally, under the SHM model, we take the only force to be opposing the direction of motion - e.g transverse - and so would surely assume T = mg...?

The answer in the back of the book is mg(1+a^2). This can quite easily be obtained from the answer I got through Conservation of Energy (mg(3-2cos(a)), using the approximation cosa = 1 - 0.5a^2 for small a. However this isn't using SHM in any form ...

So in short - am I misunderstanding what we mean when we treat a pendulum as SHM (so we DO have a radial component...?), and thus is there some method to get this answer which uses SHM more directly?

Sorry for my poor explanation - if my workings so far are unclear I can scan them in.

EDIT: Thinking about it, I see why T = mg doesn't work. We're not saying the displacement of the bob obeys SHM, but our angle made with the vertical, which is why radial acceleration doesn't have to = 0. This leads me to believe that perhaps the point of the exercise WAS just to use the cos small angle approcimation - but this isn't SHM?!

0

reply

Report

#2

Do you understand how you get the SHM approximation eqaution? Because that will answer your question. DO you know a more general form of that equation as well?

0

reply

Report

#3

actually there is a much easier way. The SHM approximation is simply that sin(theta) is approximately theta, or cos(theta) is approximately 1 - 0.5(theta)^2. Therefore you have used the SHM approximation in your argument to get the right answer.

0

reply

(Original post by

actually there is a much easier way. The SHM approximation is simply that sin(theta) is approximately theta, or cos(theta) is approximately 1 - 0.5(theta)^2. Therefore you have used the SHM approximation in your argument to get the right answer.

**Darth_Narwhale**)actually there is a much easier way. The SHM approximation is simply that sin(theta) is approximately theta, or cos(theta) is approximately 1 - 0.5(theta)^2. Therefore you have used the SHM approximation in your argument to get the right answer.

0

reply

(Original post by

I understand that if you resolve in the transverse direction (or take an energy equation and differentiate) you get theta = -(g/l)*sin(theta), and use sin(theta) is approximately theta for small theta.. However, this in itself has nothing to do with T, which only has a component in the radial direction. For the initial section I applied the usual method which is taking a conservation of energy equation and using this in mv^2/r to find an expression for T. I understand that using 1 - 0.5(theta)^2 will work - however, I don't see why this is an "SHM approximation" since I've not used the SHM equation as part of this approximation(e.g getting it into acceleration = -k*displacement via some approximation). Does this mean the question is just poorly worded or am I missing something?

**MaffsIsFun**)I understand that if you resolve in the transverse direction (or take an energy equation and differentiate) you get theta = -(g/l)*sin(theta), and use sin(theta) is approximately theta for small theta.. However, this in itself has nothing to do with T, which only has a component in the radial direction. For the initial section I applied the usual method which is taking a conservation of energy equation and using this in mv^2/r to find an expression for T. I understand that using 1 - 0.5(theta)^2 will work - however, I don't see why this is an "SHM approximation" since I've not used the SHM equation as part of this approximation(e.g getting it into acceleration = -k*displacement via some approximation). Does this mean the question is just poorly worded or am I missing something?

0

reply

Report

#6

(Original post by

e.g I understand the link between SHM and a pendulum with a small angular amplitude where you resolve in the transverse direction and use a small angle approximation to get the SHM differential equation. I also understand how to get the given answer using the small angle approximation for cosine. However, I don't see where SHM has actually come in to this particular question...

**MaffsIsFun**)e.g I understand the link between SHM and a pendulum with a small angular amplitude where you resolve in the transverse direction and use a small angle approximation to get the SHM differential equation. I also understand how to get the given answer using the small angle approximation for cosine. However, I don't see where SHM has actually come in to this particular question...

0

reply

(Original post by

So it is all to do with potential energy wells. If you imagine the PE plotted as a graph of PE against theta,, you'll obviously have a minimum at theta=0. Anything in any sort of potential energy well will undergo SHM as long as the perturbations from equilibrium are small enough that it can be modelled as a quadratic minimum. You achieve this in this instance by making the amplitude of oscillation small enough that the sine(theta) term is almost equivalent to theta. This means that we are only looking at the small section of the PE well that can be modelled as quadratic, and hence SHM will occur. By making the small angle approximation, you are satisfying the conditions for SHM. I probably haven't explained this very clearly, but do you understand what I'm getting at?

**Darth_Narwhale**)So it is all to do with potential energy wells. If you imagine the PE plotted as a graph of PE against theta,, you'll obviously have a minimum at theta=0. Anything in any sort of potential energy well will undergo SHM as long as the perturbations from equilibrium are small enough that it can be modelled as a quadratic minimum. You achieve this in this instance by making the amplitude of oscillation small enough that the sine(theta) term is almost equivalent to theta. This means that we are only looking at the small section of the PE well that can be modelled as quadratic, and hence SHM will occur. By making the small angle approximation, you are satisfying the conditions for SHM. I probably haven't explained this very clearly, but do you understand what I'm getting at?

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top