[Calc] Sign Convention in damping spring system

[WeiShan Ng]

Homework Statement

A carriage is mounted on a spring, as shown in the diagram.

The bottom of the spring is fixed to the ground. The carriage (loaded with its passenger) has a mass of 150kg. The carriage can only move vertically. The natural length of the spring is 10m and its spring constant is $k=600Nm^{-1}$. At equilibrium the spring is compressed by a distance $s$ metres. Air resistance acts on the carriage with a drag coefficient $300Nsm^{-1}$. Assume that the gravitational constant is $g=10ms^-2$.
The ride begins with the carriage released from rest 0.5m above ground level.

Let $y(t)$ be the height of the carriage below its equilibrium position at time $t$ seconds after the ride begoins. Take the positive direction to be downwards.

Homework Equations

Show that the equation of motion for this system is
$$150y''+300y'+600y=0$$

The Attempt at a Solution

$\vec{T}$ is the spring force, acting upwards, and $\vec{R}$ is the air resistive force, acting downwards, opposite to the moving direction of the carriage.
So the equation will be:
$$m\ddot{y}=mg - \vec{T} + \vec{R}$$
$$m\ddot{y} = mg - k(s+y) + \beta \dot{y}$$
$$m\ddot{y} = -ky + \beta \dot{y}$$
Since $ks=mg$
And I get $$150\ddot{y}-300\dot{y}+600y = 0$$
which have a negative coefficient for $\dot{y}$. I've double check with all the directions of the forces, and pretty sure this is the case, what have I done wrong?

 

[tnich]

Homework Statement

A carriage is mounted on a spring, as shown in the diagram.
View attachment 225636
The bottom of the spring is fixed to the ground. The carriage (loaded with its passenger) has a mass of 150kg. The carriage can only move vertically. The natural length of the spring is 10m and its spring constant is $k=600Nm^{-1}$. At equilibrium the spring is compressed by a distance $s$ metres. Air resistance acts on the carriage with a drag coefficient $300Nsm^{-1}$. Assume that the gravitational constant is $g=10ms^-2$.
The ride begins with the carriage released from rest 0.5m above ground level.

Let $y(t)$ be the height of the carriage below its equilibrium position at time $t$ seconds after the ride begoins. Take the positive direction to be downwards.

Homework Equations

Show that the equation of motion for this system is
$$150y''+300y'+600y=0$$

The Attempt at a Solution

View attachment 225639
$\vec{T}$ is the spring force, acting upwards, and $\vec{R}$ is the air resistive force, acting downwards, opposite to the moving direction of the carriage.
So the equation will be:
$$m\ddot{y}=mg - \vec{T} + \vec{R}$$
$$m\ddot{y} = mg - k(s+y) + \beta \dot{y}$$
$$m\ddot{y} = -ky + \beta \dot{y}$$
Since $ks=mg$
And I get $$150\ddot{y}-300\dot{y}+600y = 0$$
which have a negative coefficient for $\dot{y}$. I've double check with all the directions of the forces, and pretty sure this is the case, what have I done wrong?

You say that drag is acting downwards. This is not the case. It acts in the direction opposite to the velocity. If $y$, $\dot y$, and $\ddot y$ are all positive in the upward direction, then the force of drag must be negative in the upward direction. You have not written it that way in your equation for net force. I think your problem is that you have chosen the positive directions for $y$, $\dot y$, and $\ddot y$ in an inconsistent manner.

 

[WeiShan Ng]

You say that drag is acting downwards. This is not the case. It acts in the direction opposite to the velocity. If $y$, $\dot y$, and $\ddot y$ are all positive in the upward direction, then the force of drag must be negative in the upward direction. You have not written it that way in your equation for net force. I think your problem is that you have chosen the positive directions for $y$, $\dot y$, and $\ddot y$ in an inconsistent manner.

Oh I think I know where I get it wrong! So if I define the resistive force $\vec{R}$, the tension from the spring $\vec{T}$, and the weight $\vec{W}$:
$$\vec{R} = -\beta \dot{y} \hat{y} \\ \vec{T} = -ks \hat{y} - ky(t) \hat{y} \\ \vec{W} = mg \hat{y}$$
Then I use vector addition to find the net force， $\vec{F}$:
$$\vec{F}=\vec{W}+\vec{T}+\vec{R}$$
Substitute in:
$$m\ddot{y} \hat{y} = mg \hat{y} -ks\hat{y}-k(y(t))\hat{y}-\beta \dot{y} \hat{y}$$
$$m\ddot{y} \hat{y} + k(y(t)) \hat{y} + \beta \dot{y} \hat{y} = 0$$

Equating the coefficients:
$$m\ddot{y} +k(y(t))+\beta \dot{y}=0$$ And this is the equation I will be using.

So in my case I have to consider the direction of $\vec{\ddot{y}}, \vec{\dot{y}}, \vec{y}$, if they are opposite to the direction of the unit vector $\hat{y}$, then they will have a negative magnitude, so I will be subst. in a negative value into my equation; otherwise I will be substitude in a positive value.

 

[tnich]

Oh I think I know where I get it wrong! So if I define the resistive force $\vec{R}$, the tension from the spring $\vec{T}$, and the weight $\vec{W}$:
$$\vec{R} = -\beta \dot{y} \hat{y} \\ \vec{T} = -ks \hat{y} - ky(t) \hat{y} \\ \vec{W} = mg \hat{y}$$
Then I use vector addition to find the net force， $\vec{F}$:
$$\vec{F}=\vec{W}+\vec{T}+\vec{R}$$
Substitute in:
$$m\ddot{y} \hat{y} = mg \hat{y} -ks\hat{y}-k(y(t))\hat{y}-\beta \dot{y} \hat{y}$$
$$m\ddot{y} \hat{y} + k(y(t)) \hat{y} + \beta \dot{y} \hat{y} = 0$$

Equating the coefficients:
$$m\ddot{y} +k(y(t))+\beta \dot{y}=0$$ And this is the equation I will be using.

So in my case I have to consider the direction of $\vec{\ddot{y}}, \vec{\dot{y}}, \vec{y}$, if they are opposite to the direction of the unit vector $\hat{y}$, then they will have a negative magnitude, so I will be subst. in a negative value into my equation; otherwise I will be substitude in a positive value.

That is the right idea, but you are still applying it incorrectly. In which direction does the force of gravity act? When you write your equilibrium equation relating mg and ks, in which direction does the force of the spring act?

 

[tnich]

That is the right idea, but you are still applying it incorrectly. In which direction does the force of gravity act? When you write your equilibrium equation relating mg and ks, in which direction does the force of the spring act?

Oh, wait, no it is not quite the right idea. When you differentiate $y$ with respect to time, if the motion is in the positive $y$ direction, then $\dot y$ will be positive. So the $y$ and $\dot y$ vectors are positive in the same direction. The same is true of the $\dot y$ and $\ddot y$ vectors. So if you choose a positive direction for $y$, then its derivatives will be positive in that same direction.

 

[WeiShan Ng]

Oh, wait, no it is not quite the right idea. When you differentiate $y$ with respect to time, if the motion is in the positive $y$ direction, then $\dot y$ will be positive. So the $y$ and $\dot y$ vectors are positive in the same direction. The same is true of the $\dot y$ and $\ddot y$ vectors. So if you choose a positive direction for $y$, then its derivatives will be positive in that same direction.

I don't think $y$ and $\dot y$ would be necessarily in the same direction, as we can have negative velocity at the instant we are at positive coordinate. This applies to $\dot y$ and $\ddot y$ too.

 

[tnich]

I don't think $y$ and $\dot y$ would be necessarily in the same direction, as we can have negative velocity at the instant we are at positive coordinate. This applies to $\dot y$ and $\ddot y$ too.

Yes, that is true. But when they are positive, they all point in the same direction.

 

[tnich]

Yes, that is true. But when they are positive, they all point in the same direction.

To show this, start with $y\hat y$ (where $\hat y$ is the unit vector in the $y$ direction) and differentiate it twice. You get
$\dot y\hat y$ and
$\ddot y\hat y$

When $y$, $\dot y$ and $\ddot y$ are positive, in which direction do the resulting vectors point?

 

[WeiShan Ng]

To show this, start with $y\hat y$ (where $\hat y$ is the unit vector in the $y$ direction) and differentiate it twice. You get
$\dot y\hat y$ and
$\ddot y\hat y$

When $y$, $\dot y$ and $\ddot y$ are positive, in which direction do the resulting vectors point?

I don't think $y$ and $\ddot y$ will be in the same direction, as this is a damped oscillator. The direction of $\vec{T}$ will be depending on the magnitude and direction of the $y(t)$, if $y(t)$ is acting downwards, then $\vec{T}$ will always be acting downwards. If $y(t)$ is acting upwards, then we have to look at its magnitude, if $|y(t)|>s$, $\vec{T}$ will be acting upwards, if $|y(t)|<s$, $\vec{T}$ will be acting downwards. On the other hand $\vec{R}$ will always be in the opposite direction of $\dot y$. So if $\dot y$ is in positive direction, then $\vec{R}$ will be in negative direction and vice versa. And $\vec{W}$ will always be in positive direction.

So I can only start drawing my free body diagram after I account for the directions of all of my force components. This is why my first approach was wrong, I kept assuming $\vec T$ will be pointing upwards, and $\vec R$ will be pointing downwards!

I put downwards as my positive direction.

 

[tnich]

I don't think $y$ and $\ddot y$ will be in the same direction, as this is a damped oscillator.

If you mean that when $y$ is positive, $\ddot y$ is negative, that is true, but only if you pick the same direction as positive for both of them.

The direction of $\vec{T}$ will be depending on the magnitude and direction of the $y(t)$, if $y(t)$ is acting downwards, then $\vec{T}$ will always be acting downwards. If $y(t)$ is acting upwards, then we have to look at its magnitude, if $|y(t)|>s$, $\vec{T}$ will be acting upwards, if $|y(t)|<s$, $\vec{T}$ will be acting downwards. On the other hand $\vec{R}$ will always be in the opposite direction of $\dot y$. So if $\dot y$ is in positive direction, then $\vec{R}$ will be in negative direction and vice versa. And $\vec{W}$ will always be in positive direction.

It sounds like you may have figured it out here. You pick a positive direction and determine the signs relative to it.