The Student Room Group

SHM maths

Not sure how to understand this, since the particle is undergoing damped shm isn't the acceleration in the opposite direction of it's displacement by a=-w^2x however in my diagram I took the acceleration to be positiveScreenshot 2022-05-23 at 20.55.20.pngIMG_A9180DE09761-1.jpeg
(edited 1 year ago)

Scroll to see replies

What is the calculation youre doing? In undamped, unforced shm, the acceleration is as you describe. Here there is both a damping and a forcing term, so what are you asking? Also if P is displaced from O, the positive displacement arrow should be reversed?
Original post by mqb2766
What is the calculation youre doing? In undamped, unforced shm, the acceleration is as you describe. Here there is both a damping and a forcing term, so what are you asking? Also if P is displaced from O, the positive displacement arrow should be reversed?


I'm doing ma= 12sin(t) -6v -7.5x
My apologies with the poor questioning, from what you just in a damped system the acceleration is no longer always opposite to displacement?

Is the way I thought about it correct, since it's accelerating forwards I just used the f=ma resultant force to create the second order differential?
(edited 1 year ago)
Original post by Student 999
I'm doing ma= 12sin(t) -6v -7.5x
My apologies with the poor questioning, from what you just in a damped system the acceleration is no longer always opposite to displacement?

Is the way I thought about it correct, since it's accelerating forwards I just used the f=ma resultant force to create the second order differential?

Firstly your right hand side contains three variables: time, displacement and velocity. Cant see how you can easily relate them together at this point? So temporarily forgetting about the forcing term 8sin(t), then the acceleration is a function of velocity and displacement. Its a negative function of displacement to this is undamped SHM. However, the acceleration is "reduced" by the velocity which is effectively the damping part. To understand it better, youd really have to plot the spiral evolution in displacement,velocity space, but intuitively, if P has positive displacement and is moving away from O, then the acceleration is more negative to return to O so the peak displacement will be less than undamped SHM. Similarly for the other sign quadrants.
Original post by mqb2766
Firstly your right hand side contains three variables: time, displacement and velocity. Cant see how you can easily relate them together at this point? So temporarily forgetting about the forcing term 8sin(t), then the acceleration is a function of velocity and displacement. Its a negative function of displacement to this is undamped SHM. However, the acceleration is "reduced" by the velocity which is effectively the damping part. To understand it better, youd really have to plot the spiral evolution in displacement,velocity space, but intuitively, if P has positive displacement and is moving away from O, then the acceleration is more negative to return to O so the peak displacement will be less than undamped SHM. Similarly for the other sign quadrants.

Aren't the three variables just forces. Why can't you relate them?

There's a driving force of 12sin(t)
A restoring force trying to bring the particle back to the origin of 7.5x and resistant force of 6v
using these will give you the resultant force?

isn't acceleration a function of velocity against time rather than displacement.
I'm struggling to follow with the rest of the explanation could you simplify it down a bit thanks a lot
Original post by Student 999
Aren't the three variables just forces. Why can't you relate them?

There's a driving force of 12sin(t)
A restoring force trying to bring the particle back to the origin of 7.5x and resistant force of 6v
using these will give you the resultant force?

isn't acceleration a function of velocity against time rather than displacement.
I'm struggling to follow with the rest of the explanation could you simplify it down a bit thanks a lot


Yes you can relate them and form the ode as you say. Theyre forces and they sum to give the resultant and its just Newton 2.

That explanation is about as simple as you can go. There are three variables: time, displacement and velocity and these combine to give acceleration. Forget about the forcing term (time) then you just have
acceleration = - displacement - velocity
If you think about the displacement number line (so positive and negative regions), then think about the positive and negative directions of motion (velocity), you should be able to argue that the acceleration produces a more damped response. The usual displacement - time response looks like
https://beltoforion.de/en/harmonic_oscillator/
If you plot displacement against velocity, you get a spiral.
Original post by mqb2766
Yes you can relate them and form the ode as you say. Theyre forces and they sum to give the resultant and its just Newton 2.

That explanation is about as simple as you can go. There are three variables: time, displacement and velocity and these combine to give acceleration. Forget about the forcing term (time) then you just have
acceleration = - displacement - velocity
If you think about the displacement number line (so positive and negative regions), then think about the positive and negative directions of motion (velocity), you should be able to argue that the acceleration produces a more damped response. The usual displacement - time response looks like
https://beltoforion.de/en/harmonic_oscillator/
If you plot displacement against velocity, you get a spiral.

Thanks for the article/website it did confirm what I was thinkingIMG_BD8B075520D7-1.jpeg

In this question using the same approach

The acceleration is positive going upwards
Tension is also positive since it's been stretched and want to return to equilibrium
Weight is a resistance force acting against the upwards motion
14v is negative also it resists the motion of returning to equilibrium

ma=T -14v -mg
why is this wrong?
Original post by Student 999
Thanks for the article/website it did confirm what I was thinkingIMG_BD8B075520D7-1.jpeg

In this question using the same approach

The acceleration is positive going upwards
Tension is also positive since it's been stretched and want to return to equilibrium
Weight is a resistance force acting against the upwards motion
14v is negative also it resists the motion of returning to equilibrium

ma=T -14v -mg
why is this wrong?


Part a) should help. You know the vertical equilibrium length and it will be damped shm about that point. Then include the constant equilibrium length, which I guess is where the 14 on the right comes from. Note weight is not a resistance force acting against motion.
Original post by mqb2766
Part a) should help. You know the vertical equilibrium length and it will be damped shm about that point. Then include the constant equilibrium length, which I guess is where the 14 on the right comes from. Note weight is not a resistance force acting against motion.


Tension is just 24(x-0.35), why is weight not a resistance force against motion?
Original post by Student 999
Tension is just 24(x-0.35), why is weight not a resistance force against motion?


What do you think a resistance force against motion means? Why does it apply to weight/gravity?
Original post by mqb2766
What do you think a resistance force against motion means? Why does it apply to weight/gravity?

Something that requires work done which is directly involved in a system. Similar to a previous thread about circular motion about a point, weight acts as a force involved in the circular motion
Original post by Student 999
Something that requires work done which is directly involved in a system. Similar to a previous thread about circular motion about a point, weight acts as a force involved in the circular motion


Weight is a constant force. I cant see how it can be interpreted as anything to do with resistance against motion or circular motion.
Original post by mqb2766
Weight is a constant force. I cant see how it can be interpreted as anything to do with resistance against motion or circular motion.


I'm a bit confused, could you set up the equation and correct what parts I've interpreted wrong
Original post by Student 999
Not sure how to understand this, since the particle is undergoing damped shm isn't the acceleration in the opposite direction of it's displacement by a=-w^2x however in my diagram I took the acceleration to be positiveScreenshot 2022-05-23 at 20.55.20.pngIMG_A9180DE09761-1.jpeg

When you use calculus notation x is measured in the direction from 0 to P, not as you have drawn it as a 2 headed arrow. The acceleration is written d^2x/dt^2. This doesn't mean the particle is accelerating (getting faster) in the x direction because you are forgetting that this 2nd derivative can be a positive or negative number. If it is a positive number the particle is getting faster in x direction. If it is a negative number it means the particle is getting slower in the x direction.

The solution to (a) is correct with md^2x/dt^2 on the right hand side.
(edited 1 year ago)
Original post by AlanT12
When you use calculus notation x is measured in the direction from 0 to P, not as you have drawn it as a 2 headed arrow. The acceleration is written d^2x/dt^2. This doesn't mean the particle is accelerating (getting faster) in the x direction because you are forgetting that this 2nd derivative can be a positive or negative number. If it is a positive number the particle is getting faster in x direction. If it is a negative number it means the particle is getting slower in the x direction.

The solution to (a) is correct with md^2x/dt^2 on the right hand side.


I'm referring to the question I posted later about the container and spring?

My bad I thought you were mqb
(edited 1 year ago)
Original post by Student 999
I'm referring to the question I posted later about the container and spring?


In that question x is in the downwards direction from the top of the spring. The acceleration of the ball is always md^2x/dt^2. Again this can be positive or negative, so you don't add an extra sign for it. So equation of motion is

Net force in x direction = md^2x/dt^2
or
-kx+mg-14V=md^2x/dt^2
Original post by AlanT12
In that question x is in the downwards direction from the top of the spring. The acceleration of the ball is always md^2x/dt^2. Again this can be positive or negative, so you don't add an extra sign for it. So equation of motion is

Net force in x direction = md^2x/dt^2
or
-kx+mg-14V=md^2x/dt^2


ON your LHS can you explain why you took mg as positive and tension as negative from my interpretation earlier I did the opposite
Original post by Student 999
ON your LHS can you explain why you took mg as positive and tension as negative from my interpretation earlier I did the opposite

You have to remember x has a sign and direction. It is measured downwards.

The force of gravity is always downwards (in the +x direction) so is mg.

The tension in the spring is upwards for x>0 so it is in the opposite direction to x and you have to write it as -kx to show this.

.
As above, when you have a positive (downwards) displacement.

For part a), in equilibrium, the spring tension must act upwards as gravity acts downwards so the resultant force is zero
mg + T = 0
mg is the load which is acting to extend the spring so in equilibrium it must be 24e, where the extension e=x-0.35. However, the tension in the spring (force pair, Newton 3) is -24e. Note that gravity is a constant force and for a vertical spring, its effect is to shift the equilibrium position downwards from the natural length to the position mg/24+0.35. It does not affect the dynamics otherwise.

For part b) as described above, the resultant force on P (positive downwards) must be
F = mx'' = mg - 24(x-0.35) - 14x'
gravity acts to extend x downwards, the tension in the spring (for a positive extension) pulls upwards and the resistance acts against the velocity as a re-tarding force. So just rearrange. Note when x=1.167, then mg-24(1.167-0.35) = 0 as youd expect as its the equilibrium extension.

The 14 on the right hand side of the final ODE corresponds g+24*0.35/m. You could have written ithe ODE as
z'' + 7z' + 12z = 0
where z = x-14 = x-(g+24*0.35/m). Here gravity has no effect on the dynamics and the variable z is zero at the equilibrium length calculated in part a (which is a vertical shift due to gravity). So z would correspond to the extension about the spring position 1.167.
(edited 1 year ago)
Original post by AlanT12
You have to remember x has a sign and direction. It is measured downwards.

The force of gravity is always downwards (in the +x direction) so is mg.

The tension in the spring is upwards for x>0 so it is in the opposite direction to x and you have to write it as -kx to show this.

.

Thanks I didn't notice in the question displacement is taken as positive downwards

Quick Reply