# M3: SHM Watch

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why is it accelerating away from the equilibrium in the mark scheme? I thought it was suppose to accelerate towards the equilibrium point?

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#2

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why is it accelerating away from the equilibrium in the mark scheme? I thought it was suppose to accelerate towards the equilibrium point?

**Maths&physics**)why is it accelerating away from the equilibrium in the mark scheme? I thought it was suppose to accelerate towards the equilibrium point?

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Its not.

**Shaanv**)Its not.

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#4

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the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.

**Maths&physics**)the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.

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My bad, at equilibrium u would expect no acceleration so diagram a bit misleading. If particle is displaced from equilibrium then particle accelerates towards equilibrium position.

**Shaanv**)My bad, at equilibrium u would expect no acceleration so diagram a bit misleading. If particle is displaced from equilibrium then particle accelerates towards equilibrium position.

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#6

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the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.

**Maths&physics**)the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.

**upwards**from the equilibrium point, which means must also be measured in the

**upwards**direction.

But since , when x is positive, is negative so acts in the other direction.

The extension is a-x because x=0 corresponds to the equilibrium position (when the extension is a).

Notes: To show SHM, you pretty much always want to define x relative to the equilibrium position (so x = 0 at equilibrium).

The

**direction**you choose for x doesn't really matter (*), but you need to be consistent (so are taken in the same direction).

(*) unless the question actually defines it to be in a certain direction, then you need to humor the examiner.

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The diagram isn't the clearest I've ever seen, but they are taking x as the distance

But since , when x is positive, is negative so acts in the other direction.

The extension is a-x because x=0 corresponds to the equilibrium position (when the extension is a).

Notes: To show SHM, you pretty much always want to define x relative to the equilibrium position (so x = 0 at equilibrium).

The

(*) unless the question actually defines it to be in a certain direction, then you need to humor the examiner.

**DFranklin**)The diagram isn't the clearest I've ever seen, but they are taking x as the distance

**upwards**from the equilibrium point, which means must also be measured in the**upwards**direction.But since , when x is positive, is negative so acts in the other direction.

The extension is a-x because x=0 corresponds to the equilibrium position (when the extension is a).

Notes: To show SHM, you pretty much always want to define x relative to the equilibrium position (so x = 0 at equilibrium).

The

**direction**you choose for x doesn't really matter (*), but you need to be consistent (so are taken in the same direction).(*) unless the question actually defines it to be in a certain direction, then you need to humor the examiner.

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#8

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why have they taken the extension/distance between the particle and the equilibrium point, as: a - x. shouldn't it just be x as in the extension, which we sub into hooks law?

**Maths&physics**)why have they taken the extension/distance between the particle and the equilibrium point, as: a - x. shouldn't it just be x as in the extension, which we sub into hooks law?

**equilibrium**point (as opposed to the point of zero extension).

Whatever point we choose to measure x relative to, at the equilibrium point, there is no acceleration, so . But since we're looking for an expression of the form (for some constant K), we know we're not going to get anywhere unless x = 0 at the equilibrium point. So that's the point we choose.

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No. As I said in the previous post, x is the distance from the

Whatever point we choose to measure x relative to, at the equilibrium point, there is no acceleration, so . But since we're looking for an expression of the form (for some constant K), we know we're not going to get anywhere unless x = 0 at the equilibrium point. So that's the point we choose.

**DFranklin**)No. As I said in the previous post, x is the distance from the

**equilibrium**point (as opposed to the point of zero extension).Whatever point we choose to measure x relative to, at the equilibrium point, there is no acceleration, so . But since we're looking for an expression of the form (for some constant K), we know we're not going to get anywhere unless x = 0 at the equilibrium point. So that's the point we choose.

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#10

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but why are we using hookes law if we're not using the extension but the distance from the equilibrium?

**Maths&physics**)but why are we using hookes law if we're not using the extension but the distance from the equilibrium?

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You are using the extension. The extension is a-x.

**DFranklin**)You are using the extension. The extension is a-x.

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#12

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a, x would be the extension for hookes law and the distance from eq would be a - x above equilibrium... but we only use x in hookes law.

**Maths&physics**)a, x would be the extension for hookes law and the distance from eq would be a - x above equilibrium... but we only use x in hookes law.

"x" is just a letter.

Hooke's law (as you are thinking of it) says F = kx, where x is the extension.

Well, if the extension is (a-x), then Hooke's law says F = k(a-x).

And if the extension was (a-x)+3-, then Hooke's law would say

**There is nothing that says you need to use "x" in Hooke's law**.

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Regarding "We only use x in hooke's law".

"x" is just a letter.

Hooke's law (as you are thinking of it) says F = kx, where x is the extension.

Well, if the extension is (a-x), then Hooke's law says F = k(a-x).

And if the extension was (a-x)+3-, then Hooke's law would say

**DFranklin**)Regarding "We only use x in hooke's law".

"x" is just a letter.

Hooke's law (as you are thinking of it) says F = kx, where x is the extension.

Well, if the extension is (a-x), then Hooke's law says F = k(a-x).

And if the extension was (a-x)+3-, then Hooke's law would say

**There is nothing that says you need to use "x" in Hooke's law**.
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#14

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the x is suppose to represent the distance from the natural length

**Maths&physics**)the x is suppose to represent the distance from the natural length

**third time**. The solution is taking x to be the distance (upwards) from the equilibrium position.

You seem to believe that x

**must**represent the extension, but you are wrong. x is just a letter. The examiners could write a question where the particle has mass x. Are you going to insist that in this case the mass must actually be the extension?

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No. For the

You seem to believe that x

**DFranklin**)No. For the

**third time**. The solution is taking x to be the distance (upwards) from the equilibrium position.You seem to believe that x

**must**represent the extension, but you are wrong. x is just a letter. The examiners could write a question where the particle has mass x. Are you going to insist that in this case the mass must actually be the extension?sorry about that! the diagram confused me.

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**DFranklin**)

No. For the

**third time**. The solution is taking x to be the distance (upwards) from the equilibrium position.

You seem to believe that x

**must**represent the extension, but you are wrong. x is just a letter. The examiners could write a question where the particle has mass x. Are you going to insist that in this case the mass must actually be the extension?

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#17

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also, that answer for the period seems to be T = 2pi(w) as opposed to T = 2pi/w....?

**Maths&physics**)also, that answer for the period seems to be T = 2pi(w) as opposed to T = 2pi/w....?

You really need to spend more time reading the solution as opposed to jumping to conclusions.

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No it doesn't.

**DFranklin**)No it doesn't.

**You really need to spend more time reading the solution as opposed to jumping to conclusions**.
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