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Significance of the minus in M3 gravity/shm equations

I keep getting confused with why the force/acceleration equations always involve a -. Eg often used one is
mgR^2/x^2 = - ma

Why is the - there? The book says because "the acceleration is always in the direction of x decreasing". And I guess it makes sense that x is always decreasing towards the centre of Earth and force and acceleration are always toward the centre of the Earth. But I don't really get how "acceleration is always in the direction of x decreasing" relates and results to the minus in the equation. I'm especially getting confused when it's objects falling toward the surface. In this case the speed is increasing in the direction of the net force, why is a minus necessary? So what if it's in the direction of x decreasing?
I could ask this same question about a simple pendulum. "The restoring force is always in the direction of x decreasing" Why does that result in a minus being in the equation?
I hope I made sense, any help is much appreciated.
That particular equation I'm not sure about. How Ive been taught is that the acceleration in SHM is always directed opposite to the displacement. So, like with a pendulum as it's going towards the centre, there is still an acceleration that will be acting to put it back to where it came from. I've havent seen many gravity questions/material, so idk.
Reply 2
Original post by mrno1324
I keep getting confused with why the force/acceleration equations always involve a -. Eg often used one is
mgR^2/x^2 = - ma

Why is the - there? The book says because "the acceleration is always in the direction of x decreasing". And I guess it makes sense that x is always decreasing towards the centre of Earth and force and acceleration are always toward the centre of the Earth. But I don't really get how "acceleration is always in the direction of x decreasing" relates and results to the minus in the equation. I'm especially getting confused when it's objects falling toward the surface. In this case the speed is increasing in the direction of the net force, why is a minus necessary? So what if it's in the direction of x decreasing?
I could ask this same question about a simple pendulum. "The restoring force is always in the direction of x decreasing" Why does that result in a minus being in the equation?
I hope I made sense, any help is much appreciated.



Original post by Plentycarbon
That particular equation I'm not sure about. How Ive been taught is that the acceleration in SHM is always directed opposite to the displacement. So, like with a pendulum as it's going towards the centre, there is still an acceleration that will be acting to put it back to where it came from. I've havent seen many gravity questions/material, so idk.


acceleration is the rate of change, of the rate of of change of displacement with respect to time.


In SHM
So look for the direction of x increasing first. Then this is the way the "acceleration" is pointing.. Forget about physical acceleration.


In Circular motion including gravitational problems:

Acceleration must be marked in the direction of r (radius) increasing.
But in circular motion acceleration is towards the centre so it is minus.
(this comes from the general expression of acceleration in polar coordinates)
Reply 3
Original post by TeeEm
acceleration is the rate of change, of the rate of of change of displacement with respect to time.


In SHM
So look for the direction of x increasing first. Then this is the way the "acceleration" is pointing.. Forget about physical acceleration.


In Circular motion including gravitational problems:

Acceleration must be marked in the direction of r (radius) increasing.
But in circular motion acceleration is towards the centre so it is minus.
(this comes from the general expression of acceleration in polar coordinates)

Thanks for the reply. I'm not sure I understand the difference between what you call physical acceleration and acceleration at the rate of change. Eg if we look at a simple pendulum head on while it's swinging from the equilibrium to the right and call this direction (right) positive. The displacement is increasing positively (to the right), but surely it's accelerating in the opposite direction (that the bob is slowing down).
And wouldn't the main formula of SHM a = -omega^2 x imply that acceleration is always in the opposite direction to displacement, because of the minus?
Reply 4
Original post by mrno1324
Thanks for the reply. I'm not sure I understand the difference between what you call physical acceleration and acceleration at the rate of change. Eg if we look at a simple pendulum head on while it's swinging from the equilibrium to the right and call this direction (right) positive. The displacement is increasing positively (to the right), but surely it's accelerating in the opposite direction (that the bob is slowing down).
And wouldn't the main formula of SHM a = -omega^2 x imply that acceleration is always in the opposite direction to displacement, because of the minus?


I agree that it is very subtle and you do not quite understand the difference... I had the same difficulty when I was learning and so do all of my students.

very to explain on line but I will try again.


imagine the standard SHM scenario; a particle hanging in equilibrium.

It is pulled downwards and released so its physical speed is up and its physical acceleration is also up.

NOW TRY TO FOLLOW THIS

If x is the displacement of the particle from the equilibrium position then

x=0 is at equilibrium position
x=1 is 1 metre below the equilibrium position
x=2 is 2 metres below the equilibrium position
x=3 is 3 metres below the equilibrium position
etc

so x is increasing downwards
then the mathematical velocity (rate of change of x) is downwards although the particle is physically moving up
the mathematical acceleration (2nd rate of change of x) is downwards although the particle is speeding up


Ask your teacher if any good.
Reply 5
Original post by TeeEm
I agree that it is very subtle and you do not quite understand the difference... I had the same difficulty when I was learning and so do all of my students.

very to explain on line but I will try again.


imagine the standard SHM scenario; a particle hanging in equilibrium.

It is pulled downwards and released so its physical speed is up and its physical acceleration is also up.

NOW TRY TO FOLLOW THIS

If x is the displacement of the particle from the equilibrium position then

x=0 is at equilibrium position
x=1 is 1 metre below the equilibrium position
x=2 is 2 metres below the equilibrium position
x=3 is 3 metres below the equilibrium position
etc

so x is increasing downwards
then the mathematical velocity (rate of change of x) is downwards although the particle is physically moving up
the mathematical acceleration (2nd rate of change of x) is downwards although the particle is speeding up


Ask your teacher if any good.


I think I understand it better now. Could you please check if I got this right.
So the rate of change of *something* is mathematically ALWAYS in the same direction as *something*, is that right?
When considering SHM from a physics perspective and Newton's second law acceleration is ALWAYS in the same direction as the net force, which is to the centre of oscillation (equilibrium). But from a maths perspective there is also a mathematical acceleration separate from the physical, which is in the opposite direction of the physical acceleration.
I wouldn't say all this is a minor detail, why do mathematics and physics, (that are usually interchangeable when mechanics is concerned) clash so bad and just in this specific topic of SHM?
Thanks again
Reply 6
Original post by mrno1324
I think I understand it better now. Could you please check if I got this right.
So the rate of change of *something* is mathematically ALWAYS in the same direction as *something*, is that right?


A sensible convention

Original post by mrno1324

When considering SHM from a physics perspective and Newton's second law acceleration is ALWAYS in the same direction as the net force, which is to the centre of oscillation (equilibrium). But from a maths perspective there is also a mathematical acceleration separate from the physical, which is in the opposite direction of the physical acceleration.
I wouldn't say all this is a minor detail, why do mathematics and physics, (that are usually interchangeable when mechanics is concerned) clash so bad and just in this specific topic of SHM?


A level physics is a JOKE!!!

THERE IS NO PHYSICS WITHOUT MATHS



At proper physics level there are no clashes.
The conventions are the same for mathematics and physics at degree level.
Reply 7
Original post by TeeEm
A sensible convention



A level physics is a JOKE!!!

THERE IS NO PHYSICS WITHOUT MATHS



At proper physics level there are no clashes.
The conventions are the same for mathematics and physics at degree level.


That's really helpful, I feel a lot better know, thanks a bunch.
Reply 8
Original post by mrno1324
That's really helpful, I feel a lot better know, thanks a bunch.


no worries

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