M3: SHM Watch

Maths&physics
Badges: 15
Rep:
?
#1
Report Thread starter 1 year ago
#1
why is it accelerating away from the equilibrium in the mark scheme? I thought it was suppose to accelerate towards the equilibrium point?
0
reply
Shaanv
Badges: 17
Rep:
?
#2
Report 1 year ago
#2
(Original post by 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?
Its not.
Posted on the TSR App. Download from Apple or Google Play
0
reply
Maths&physics
Badges: 15
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by Shaanv)
Its not.
the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.
0
reply
Shaanv
Badges: 17
Rep:
?
#4
Report 1 year ago
#4
(Original post by Maths&physics)
the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.
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.
Posted on the TSR App. Download from Apple or Google Play
0
reply
Maths&physics
Badges: 15
Rep:
?
#5
Report Thread starter 1 year ago
#5
(Original post by 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.
why have they done: (x-a) as the extension?
0
reply
DFranklin
Badges: 18
Rep:
?
#6
Report 1 year ago
#6
(Original post by Maths&physics)
the equilibrium is at 4a... and the arrow for acceleration seems to be pointing away.
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 \ddot{x} must also be measured in the upwards direction.

But since \ddot x = -\frac{g}{a} x, when x is positive, \ddot{x} 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 \dot{x}, \ddot{x} 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.
0
reply
Maths&physics
Badges: 15
Rep:
?
#7
Report Thread starter 1 year ago
#7
(Original post by 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 \ddot{x} must also be measured in the upwards direction.

But since \ddot x = -\frac{g}{a} x, when x is positive, \ddot{x} 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 \dot{x}, \ddot{x} 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.
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?
0
reply
DFranklin
Badges: 18
Rep:
?
#8
Report 1 year ago
#8
(Original post by 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?
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 \ddot{x} = 0. But since we're looking for an expression of the form \ddot{x} = -K x (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.
0
reply
Maths&physics
Badges: 15
Rep:
?
#9
Report Thread starter 1 year ago
#9
(Original post by 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 \ddot{x} = 0. But since we're looking for an expression of the form \ddot{x} = -K x (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.
but why are we using hookes law if we're not using the extension but the distance from the equilibrium?
0
reply
DFranklin
Badges: 18
Rep:
?
#10
Report 1 year ago
#10
(Original post by Maths&physics)
but why are we using hookes law if we're not using the extension but the distance from the equilibrium?
You are using the extension. The extension is a-x.
0
reply
Maths&physics
Badges: 15
Rep:
?
#11
Report Thread starter 1 year ago
#11
(Original post by DFranklin)
You are using the extension. The extension is a-x.
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.
Attached files
0
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 1 year ago
#12
(Original post by 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.
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-\int_0^x e^{-\xi^2}\,d\xi, then Hooke's law would say

F = k [(a-x)+3-\int_0^x e^{-\xi^2}\,d\xi]

There is nothing that says you need to use "x" in Hooke's law.
0
reply
Maths&physics
Badges: 15
Rep:
?
#13
Report Thread starter 1 year ago
#13
(Original post by 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-\int_0^x e^{-\xi^2}\,d\xi, then Hooke's law would say

F = k [(a-x)+3-\int_0^x e^{-\xi^2}\,d\xi]

There is nothing that says you need to use "x" in Hooke's law.
the x is suppose to represent the distance from the natural length and in the mark scheme, x represents that but the distance beyond the natural length is: a-x??? that would give us the distance from the equilibrium and not the distance from the natural length. im so confused. lol
0
reply
DFranklin
Badges: 18
Rep:
?
#14
Report 1 year ago
#14
(Original post by Maths&physics)
the x is suppose to represent the distance from the natural length
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?
0
reply
Maths&physics
Badges: 15
Rep:
?
#15
Report Thread starter 1 year ago
#15
(Original post by 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?
ah ok...... my bad! lol

sorry about that! the diagram confused me.
0
reply
Maths&physics
Badges: 15
Rep:
?
#16
Report Thread starter 1 year ago
#16
(Original post by 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?
also, that answer for the period seems to be T = 2pi(w) as opposed to T = 2pi/w....?
0
reply
DFranklin
Badges: 18
Rep:
?
#17
Report 1 year ago
#17
(Original post by Maths&physics)
also, that answer for the period seems to be T = 2pi(w) as opposed to T = 2pi/w....?
No it doesn't.

You really need to spend more time reading the solution as opposed to jumping to conclusions.
0
reply
Maths&physics
Badges: 15
Rep:
?
#18
Report Thread starter 1 year ago
#18
(Original post by DFranklin)
No it doesn't.

You really need to spend more time reading the solution as opposed to jumping to conclusions.
indeed, I do. yeah, so: 1/\sqrt{(a/b)} = \sqrt{b/a}
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Manchester Metropolitan University
    Undergraduate Open Day Undergraduate
    Wed, 19 Jun '19
  • University of West London
    Undergraduate Open Day - West London Campus Undergraduate
    Wed, 19 Jun '19
  • University of Warwick
    Undergraduate Open Day Undergraduate
    Fri, 21 Jun '19

How did your AQA A-level Biology Paper 3 go?

Loved the paper - Feeling positive (333)
15.6%
The paper was reasonable (1164)
54.55%
Not feeling great about that exam... (461)
21.6%
It was TERRIBLE (176)
8.25%

Watched Threads

View All