I forgot how this formula comes in the simple harmonic motion. I mean but then how do you substitute in place of r^2. Something to do with circles. Just couldn't find in the M3 book. Please help me or give me some links where i can find this formula.

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 25072009 03:55
Last edited by ssadi; 25072009 at 04:00. 
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Unbounded
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 25072009 04:06
With my little M3 knowledge, I'm wondering whether you're confused between circular motion and simple harmonic motion. We know that the acceleration is proportional to the displacement for SHM, with the opposite direction towards the origin, so we can write:
for some constant .
Using that , we get
, and noting that at the maximum amplitude, the velocity is zero, and the displacement is plus or minus the amplitude, i.e. , and hence
ThereforeLast edited by Unbounded; 25072009 at 04:53. 
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 25072009 04:16
(Original post by GHOSH5)
With my little M3 knowledge, I'm wondering whether you're confused between circular motion and simple harmonic motion. We know that the acceleration is proportional to the displacement for SHM, with the opposite direction towards the origin, so we can write:
for some constant .
Using that , we get
, and noting that at the maximum amplitude, the velocity is zero, and the displacement is plus or minus the acceleration, i.e. , and hence
Therefore
Can you make a link between the r and the a and the x. I mean a^2=x^2+r^2 i can deduce, but isn't r and a the same? I'm confused with that, because i know circular motion can be translated to simple harmonic motion, but what is the connection between amplitude and radius i cannot make out, but it seems to me that they must be equal, but when i take into consideration v=rw, you see this is bugging me. 
tazarooni89
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 25072009 04:40
(Original post by ssadi)
Thanks, it clarifies where it came from.
Can you make a link between the r and the a and the x. I mean a^2=x^2+r^2 i can deduce, but isn't r and a the same? I'm confused with that, because i know circular motion can be translated to simple harmonic motion, but what is the connection between amplitude and radius i cannot make out, but it seems to me that they must be equal, but when i take into consideration v=rw, you see this is bugging me.
Imagine a particle travelling in a circle at a constant speed, and you're looking at this circle from a side view  so to you, it looks like a particle moving back and forth along a straight line observing simple harmonic motion.
The v in v^2 = (w^2)(r^2) is the actual velocity of the particle
But the v in v^2= (w^2)(a^2  x^2) is the apparent velocity of the. particle, from the angle that you're looking at it. You're only observing a component of it's true velocity. It's probably better to rewrite this v as dx/dt.
You can't combine the two equations v^2 = (w^2)(r^2) and v^2= (w^2)(a^2  x^2), because the letter v is representing something different in each case.
Anyway, a possible deduction of the formula v^2= (w^2)(a^2  x^2) is this:
By definition, simple hamonic motion abides by the following formula: x" + (w^2)x = 0
(Where ' denotes differentiation with respect to time, x is the displacement of the particle from the centre of oscillation, and w is a constant)
If you solve this as you'd solve any second order differential equation:
Auxilliary Equation  m^2 + w^2 = 0
m = iw, iw
We then have x = Acos(wt) + Bsin(wt), where A and B are arbitrary constants.
Differentiate both sides, x' = Bwcos(wt) = Awsin(wt)
Now plug in some initial conditions that we know about simple harmonic motion
For example, when t = 0, x = a, x' = 0  (here, a denotes the amplitude of oscillation).
It follows that A = a, and B = 0
So now we have:
x = acos(wt)  (1)
x' = awsin(wt)  (2)
Square Equation (2), and we get:
(x')^2 = (a^2)(w^2)(sin^2 wt)
(x')^2 = (a^2)(w^2)(1  cos^2 wt)
(x')^2 = (w^2)(a^2  a^2 cos^2 wt)  (3)
Square Equation (1), and we get:
x^2 = a^2 cos^2 (wt) (4)
Combine equations (3) and (4), and we have
(x')^2 = (w^2)(a^2  x^2) 
Unbounded
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 25072009 04:46
(Original post by ssadi)
Thanks, it clarifies where it came from.
Can you make a link between the r and the a and the x. I mean a^2=x^2+r^2 i can deduce, but isn't r and a the same? I'm confused with that, because i know circular motion can be translated to simple harmonic motion, but what is the connection between amplitude and radius i cannot make out, but it seems to me that they must be equal, but when i take into consideration v=rw, you see this is bugging me.
Circular motion tells us about the relationship between P and O as P goes around the circle. SHM tells us about the relationship between Q and O as Q moves from A to B, and you seem to be getting a little confused between the two. From what it seems, you think that the equation v = rw represents the velocity of Q towards O, however, this is clearly not true, as it represents the velocity of P around the circle, which is clearly constant, as r and w are constant. However the v in the equation that I derived does represent the velocity of Q towards O. So we must be careful not to mix up the two different velocities! Easily done, but hopefully in future, you'll remember
As you know, we can show that the constant I used originally, turns out to be the angular velocity of P, namely omega (w). This can be done by considering the rightangled triangle PQO. We can show that the acceleration of Q towards O is equal to the acceleration of P towards O times the cosine of the angle theta (think trigonometry), or alternatively:
, where a is the acceleration of Q towards O. But with trigonometry again, we know that , and hence . But since the acceleration is directed towards O, whereas the displacement is 'going the other way' essentially, we find that , which means that my constant I originally used, , is indeed the angular velocity of P, omega (w). And so we end up with your equation:
. [But remember, this 'v' is not the same as the 'v' in the equation v = rw].
I hope that helps. 
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 25072009 05:52
(Original post by GHOSH5)
With my little M3 knowledge, I'm wondering whether you're confused between circular motion and simple harmonic motion. We know that the acceleration is proportional to the displacement for SHM, with the opposite direction towards the origin, so we can write:
for some constant .
Using that , we get
, and noting that at the maximum amplitude, the velocity is zero, and the displacement is plus or minus the amplitude, i.e. , and hence
Therefore 
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 25072009 12:31
(Original post by tazarooni89)
It's not to do with substitution, because the two v's in the two formulae you quoted aren't representing the same thing.
Imagine a particle travelling in a circle at a constant speed, and you're looking at this circle from a side view  so to you, it looks like a particle moving back and forth along a straight line observing simple harmonic motion.
The v in v^2 = (w^2)(r^2) is the actual velocity of the particle
But the v in v^2= (w^2)(a^2  x^2) is the apparent velocity of the. particle, from the angle that you're looking at it. You're only observing a component of it's true velocity. It's probably better to rewrite this v as dx/dt.
You can't combine the two equations v^2 = (w^2)(r^2) and v^2= (w^2)(a^2  x^2), because the letter v is representing something different in each case.
Anyway, a possible deduction of the formula v^2= (w^2)(a^2  x^2) is this:
By definition, simple hamonic motion abides by the following formula: x" + (w^2)x = 0
(Where ' denotes differentiation with respect to time, x is the displacement of the particle from the centre of oscillation, and w is a constant)
If you solve this as you'd solve any second order differential equation:
Auxilliary Equation  m^2 + w^2 = 0
m = iw, iw
We then have x = Acos(wt) + Bsin(wt), where A and B are arbitrary constants.
Differentiate both sides, x' = Bwcos(wt) = Awsin(wt)
Now plug in some initial conditions that we know about simple harmonic motion
For example, when t = 0, x = a, x' = 0  (here, a denotes the amplitude of oscillation).
It follows that A = a, and B = 0
So now we have:
x = acos(wt)  (1)
x' = awsin(wt)  (2)
Square Equation (2), and we get:
(x')^2 = (a^2)(w^2)(sin^2 wt)
(x')^2 = (a^2)(w^2)(1  cos^2 wt)
(x')^2 = (w^2)(a^2  a^2 cos^2 wt)  (3)
Square Equation (1), and we get:
x^2 = a^2 cos^2 (wt) (4)
Combine equations (3) and (4), and we have
(x')^2 = (w^2)(a^2  x^2) 
ssadi
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 25072009 12:32
(Original post by GHOSH5)
Hopefully this will help. I can't find a picture on the internet, but I'll describe a drawing to you (and it'll probably help you understand to draw it out anyway). Draw a circle. Draw its diameter [horizontally]. Label the centre (midpoint of diameter) as point O, and label the ends of the diameter A and B. Pick a point on the circle and label it P (a general point, where our particle is moving round). Draw a little curved arrow at P going around the circle, indicating its angular velocity, and label it omega (w) [constant]. Draw the line linking P and O  the radius, and label the length r [constant]. Now draw a point from P to a point on the diameter Q such that PQ and the diameter are perpendicular [a vertical line, given the diameter should have been drawn horizontally]. Mark the rightangle OPQ. Label the angle POQ theta. Mark the distance between O and Q as x.
Circular motion tells us about the relationship between P and O as P goes around the circle. SHM tells us about the relationship between Q and O as Q moves from A to B, and you seem to be getting a little confused between the two. From what it seems, you think that the equation v = rw represents the velocity of Q towards O, however, this is clearly not true, as it represents the velocity of P around the circle, which is clearly constant, as r and w are constant. However the v in the equation that I derived does represent the velocity of Q towards O. So we must be careful not to mix up the two different velocities! Easily done, but hopefully in future, you'll remember
As you know, we can show that the constant I used originally, turns out to be the angular velocity of P, namely omega (w). This can be done by considering the rightangled triangle PQO. We can show that the acceleration of Q towards O is equal to the acceleration of P towards O times the cosine of the angle theta (think trigonometry), or alternatively:
, where a is the acceleration of Q towards O. But with trigonometry again, we know that , and hence . But since the acceleration is directed towards O, whereas the displacement is 'going the other way' essentially, we find that , which means that my constant I originally used, , is indeed the angular velocity of P, omega (w). And so we end up with your equation:
. [But remember, this 'v' is not the same as the 'v' in the equation v = rw].
I hope that helps.
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