# M3 Further Dynamics - alternative approach to a solutionWatch

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Thread starter 4 years ago
#1
in the M3 Keith Pledger edexcel textbook, page 66 example 15:

A particle P of mass 0.2kg is attached to one end of a light elastic string of natural length 0.6m and modulus of elasticity 8N. The other end of the string is fixed to a point A on a ceiling. When the particle is hanging in equilibrium the length of the string is Lm.

a)calculate the value of L,

using Hooke's law I obtained L =0.747

The particle is held at A and released from rest. It first comes to instantaneous rest when the length of the string is Km.

b)Use Energy Considerations to calculate the value of K.

Now I know how to use the energy equations to solve the answer, and in side of the page the note to the example says "the question states that you must do this part using conservation of energy"..

Now that is all ok, but I got the message that there may be another way to solve this ==> using the geometrical methods equations
(ie v^2 = w^2 (a^2 - x^2)

I have tried using geometrical methods but have not been able to find the right answer this way. I have got the value w = 200/3.

and using the information from part b, am I right in setting v = 0 in the geometrical methods equation?

Can someone confirm if part b is solvable using geometrical methods ( if it is, could they please show me how?)

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4 years ago
#2
(Original post by Coral Reafs)

Now that is all ok, but I got the message that there may be another way to solve this ==> using the geometrical methods equations
(ie v^2 = w^2 (a^2 - x^2)
Never heard it called that before. This is one of the SHM equations. Which are valid whilst the acceleration is proportional to the displacment about some equilibrium point and in the opposite direction.

However, in this case, when the particle is released it is initially falling freely under gravity and is not undergoing SHM. So that equation won't be valid for the entire motion.
Thread starter 4 years ago
#3
(Original post by ghostwalker)
Never heard it called that before. This is one of the SHM equations. Which are valid whilst the acceleration is proportional to the displacment about some equilibrium point and in the opposite direction.

However, in this case, when the particle is released it is initially falling freely under gravity and is not undergoing SHM. So that equation won't be valid for the entire motion.
if you can sometime grab hold of the m3 book the "geometrical methods" bit is on pages 48 - 51. It just basically uses a circle and right angled triangles along with equations in sin and cos.

and thank you for clarifying it for me, its much appreciated as I'm self teaching 0
4 years ago
#4
(Original post by Coral Reafs)
if you can sometime grab hold of the m3 book the "geometrical methods" bit is on pages 48 - 51. It just basically uses a circle and right angled triangles along with equations in sin and cos.
'Fraid I don't have the book, so can't check that out.

4 years ago
#5
I've got it at school, so will try and get back to you tomorrow to explain what they've done.
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Thread starter 4 years ago
#6
(Original post by Mr T Pities You)
I've got it at school, so will try and get back to you tomorrow to explain what they've done.
ok that will be much appreciated 0
4 years ago
#7
(Original post by ghostwalker)
Never heard it called that before. This is one of the SHM equations. Which are valid whilst the acceleration is proportional to the displacment about some equilibrium point and in the opposite direction.

However, in this case, when the particle is released it is initially falling freely under gravity and is not undergoing SHM. So that equation won't be valid for the entire motion.
My ancient M3 book (2000 syllabus!) refers to the "geometric approach" to SHM whereby they just mean deriving x'' = -(w^2)x when x is the x-component of a particle moving round a circle of radius a with angular speed w. Nothing really profound!

(They also say the circle is also called the "reference circle" which isn't a term I've ever seen elsewhere!)
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4 years ago
#8
(Original post by davros)
My ancient M3 book (2000 syllabus!) refers to the "geometric approach" to SHM whereby they just mean deriving x'' = -(w^2)x when x is the x-component of a particle moving round a circle of radius a with angular speed w. Nothing really profound!

(They also say the circle is also called the "reference circle" which isn't a term I've ever seen elsewhere!)
Damn new fangled approaches I feel a "during the war" coming on - to be said in grandad's voice from "Fools and Horses".
My reference work is from the 70s. PRSOM
4 years ago
#9
(Original post by Coral Reafs)
if you can sometime grab hold of the m3 book the "geometrical methods" bit is on pages 48 - 51. It just basically uses a circle and right angled triangles along with equations in sin and cos.

and thank you for clarifying it for me, its much appreciated as I'm self teaching Just had a look at this and the book is pretty clear. One thing I would say is that you don't need much of this, and can just quote the results. When you cover second order differential equations (not in M3!) then this would make more sense. The bullet points on page 48 and 51 are sufficient. I know that's not a great answer, but this is one of those areas where a further explanation would require a face-to-face; I couldn't improve on the books written explanation.
As for the bit titled 'geometrical methods' this is just using the reference circle. You can answer a lot of questions using this, and I would say it's the best way of solving questions where you need to find time travelled between two given points. As it's a diagram, I think it's easier to use than blindly launching into a set of formulae.
This probably isn't too helpful, but there are times in maths where you have to be able to do something before you understand it!
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Thread starter 4 years ago
#10
(Original post by Mr T Pities You)
Just had a look at this and the book is pretty clear. One thing I would say is that you don't need much of this, and can just quote the results. When you cover second order differential equations (not in M3!) then this would make more sense. The bullet points on page 48 and 51 are sufficient. I know that's not a great answer, but this is one of those areas where a further explanation would require a face-to-face; I couldn't improve on the books written explanation.
As for the bit titled 'geometrical methods' this is just using the reference circle. You can answer a lot of questions using this, and I would say it's the best way of solving questions where you need to find time travelled between two given points. As it's a diagram, I think it's easier to use than blindly launching into a set of formulae.
This probably isn't too helpful, but there are times in maths where you have to be able to do something before you understand it!
ok, so you're saying aswell that in the example, part b cannot be solved using the formula concerning v,a,w and x?

p.s. I've covered 2nd order differential equations in fp2(im self teaching A2 FM - m3 fp2 fp3)
0
4 years ago
#11
(Original post by Coral Reafs)
ok, so you're saying aswell that in the example, part b cannot be solved using the formula concerning v,a,w and x?

p.s. I've covered 2nd order differential equations in fp2(im self teaching A2 FM - m3 fp2 fp3)
As Ghostwalker said above, the vawx formula is a result of SHM. You can only use it if you've got SHM. You'll notice that in the example part c gets you to show that it is SHM, so to use vawx formula you'd have to do that first!
In this question it isn't moving with SHM the whole time, so you can't use it for the whole scenario.
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