# Equation of an oscillating system without any starting values

• RiotRick
No, that is the force at time t. I mean express xmax in terms of x0, v0 and ω.xmax = x0*cos(wt)+v0*sin(wt)*ω

## Homework Statement

A mass m1 is located on a platform with mass M. The platfrom is located on springs with total constant k such that it can swing vertically in direction x.

a) Write down the equations of motion assuming mass m1 will always be connected to the platform. Write it as x(t)

b) What's the maximum force on the spring?

c) What's the maximum normal force between m1 and M?

d) What's the Amplitude so that m1 will takeoff from the platform

## Homework Equations

$$m*\frac{d^2x}{dt^2} = -kx$$

## The Attempt at a Solution

a)
I have no information given, so I assume I can't use sin() or cos() or am I overthinking it?
Let m = m1+M
So we have $$m*\frac{d^2x}{dt^2} = -kx$$
And I use the Cauchy–Euler equation $$m*\lambda^2*e^{{\lambda}*t} + k*e^{{\lambda}*t}=0$$
Which leads to: $$\lambda_{1,2} = +-\sqrt{\frac{-k}{m}}$$
Here I have a minus inside the root. I think since no direction is given I can say that I only care about the absolute value

So I have $$A*e^{\sqrt{k/m}*t}+B*e^{-\sqrt{k/m}*t}$$
There is nothing more that I can do about the constants, because there is nothing given.

For b) and c) I know it has to be on the bottom dead center. d) The acceleration of the mass on the spring has to be greater than gravity
How do I continue from here? My ideas would be to solve it for ##\sqrt{k/m}## or use ##E_{pot} + E_{kin}## but it seems I'm totally off track.

Mister T said:
Do I have to add it directly in the "relevant equation" and then treat it like inhomogeneous differential equation 2nd order? or add just afterwards?

RiotRick said:
Do I have to add it directly in the "relevant equation" and then treat it like inhomogeneous differential equation 2nd order? or add just afterwards?
It depends how you are defining x. Usualy in a vertical spring oscillation one takes x as displacement from equilibrium. This means you are only concerned with the additional spring force over and above that required to oppose g. Thus, your equation is correct.

RiotRick said:
since no direction is given I can say that I only care about the absolute value
No, the sign matters. Your lambda becomes imaginary, giving terms like e±iωt. You should recognise what that means.
The usual solution avoids this by going straight to the well known solution to ##\ddot x+kx=0##. This is SHM - what form of equation would you expect?

haruspex said:
No, the sign matters. Your lambda becomes imaginary, giving terms like e±iωt. You should recognise what that means.
The usual solution avoids this by going straight to the well known solution to ##\ddot x+kx=0##. This is SHM - what form of equation would you expect?
The Task says I have to give it as a function x(t). I can solve a to the point where x(t) = ##x_0*cos(wt) + \frac{v_0}{w}*sin(wt)## but how can I use this to solve b,c and d? We can also take Acos()

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For b)
Can I say ##mgx = \frac{1}{2}kx^2## solve it for x and insert it in ##F = -kx## which will give me ##F=2g##

RiotRick said:
The Task says I have to give it as a function x(t). I can solve a to the point where x(t) = ##x_0*cos(wt) + \frac{v_0}{w}*sin(wt)## but how can I use this to solve b,c and d? We can also take Acos()
If you have quoted the whole question as given to you, there is not enough information. The best you can do is express the answers in terms of the unknowns you have introduced.

haruspex said:
If you have quoted the whole question as given to you, there is not enough information. The best you can do is express the answers in terms of the unknowns you have introduced.
so for a) the answer is simply ##(m+M)x'' = -kx_{max}##?

RiotRick said:
so for a) the answer is simply ##(m+M)x'' = -kx_{max}##?
No, xmax is a constant surely?

haruspex said:
No, xmax is a constant surely?
I'm sorry. I'm talking about b)

RiotRick said:
I'm sorry. I'm talking about b)
Ok, but express it in terms of the unknowns you introduced in post #6.

haruspex said:
Ok, but express it in terms of the unknowns you introduced in post #6.
Do you mean ##(m+M)x'' = -k*[x_0*cos(wt)+\frac{v_0}{w}*sin(wt) ##

RiotRick said:
Do you mean ##(m+M)x'' = -k*[x_0*cos(wt)+\frac{v_0}{w}*sin(wt) ##
No, that is the force at time t. I mean express xmax in terms of x0, v0 and ω.

haruspex said:
No, that is the force at time t. I mean express xmax in terms of x0, v0 and ω.
The derivative of that function has to be 0 but that's hard to do with that function. I don't see any approach

RiotRick said:
The derivative of that function has to be 0 but that's hard to do with that function. I don't see any approach
Energy?

haruspex said:
Energy?
##mgx=1/2*kx^2## putting my function x(t) in will still leave me with cos sin stuff

RiotRick said:
##mgx=1/2*kx^2## putting my function x(t) in will still leave me with cos sin stuff
I thought we agreed g becomes irrelevant by choice of x=0 at equilibrium?
In terms of your unknowns in post #6, what is the initial energy (excluding GPE)?

haruspex said:
I thought we agreed g becomes irrelevant by choice of x=0 at equilibrium?
In terms of your unknowns in post #6, what is the initial energy (excluding GPE)?
the initial energy would be ##x_0## and ##v_0##

RiotRick said:
the initial energy would be ##x_0## and ##v_0##
It would be determined by those, yes, but what is the algebraic expression?
What about when the spring force is maximised?

haruspex said:
It would be determined by those, yes, but what is the algebraic expression?
What about when the spring force is maximised?
Sorry I don't see it

RiotRick said:
Sorry I don't see it
This what I had in mind:
In terms of x and v, what is the expression for total mechanical energy (remember, we have neutralised GPE) at any time t?
In terms of the unknowns you introduced in post #6, what is its value at t=0?
In terms of x and v, what is its value when the spring force is at maximum?

However, it was only a suggested approach and I have not checked that it helps. If you prefer, you can wait until I have had a chance to try it myself.

Edit: see next post

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it turns out that using energy leads to the same place, but that place is not as bad as you made out. You just need to use sec2=1+tan2.
Please post your attempt at (b) as far as you can get.

haruspex said:
it turns out that using energy leads to the same place, but that place is not as bad as you made out. You just need to use sec2=1+tan2.
Please post your attempt at (b) as far as you can get.
I guess you want me to use one of those magic identities. The only one I know is ##a*sin(x) + b*cos(x) = sqrt(a^2 + b^2)*sin(x + t)## so I have ##\sqrt{(x_0)^2+(\frac{v_0}{w})^2}##

So I can read out the Amplitude

For C)

It leads me to ## \frac{F}{m+M} = \sqrt{(x_0)^2+(\frac{v_0}{w})^2}k##
Does this look reasonable?

RiotRick said:
It leads me to ## \frac{F}{m+M} = \sqrt{(x_0)^2+(\frac{v_0}{w})^2}k##
Does this look reasonable?
Close, but you have a dimensional inconsistency (always something worth checking!)
The LHS is an acceleration but the right is a force.

haruspex said:
Close, but you have a dimensional inconsistency (always something worth checking!)
The LHS is an acceleration but the right is a force.
Ok step by step.
##\frac{F}{m+M}= x'' = \frac{-kx}{m+M}##
##F_n = \frac{mF}{m+M}= -kx##
##\frac{F}{m+M} = \sqrt{(x_0)^2+(\frac{v_0}{w})^2}\frac{k}{m}##

RiotRick said:
Ok step by step.
##\frac{F}{m+M}= x'' = \frac{-kx}{m+M}##
##F_n = \frac{mF}{m+M}= -kx##
##\frac{F}{m+M} = \sqrt{(x_0)^2+(\frac{v_0}{w})^2}\frac{k}{m}##
That's got the dimensionality right, but for this part of the question the two masses might as well be one, so you should have M+m both sides, which then cancels.

RiotRick
When I look at this:

RiotRick said:
a) Write down the equations of motion assuming mass m1 will always be connected to the platform. Write it as x(t)

I'm thinking they want two equations. One for ##M## and one for ##m_1##.

Mister T said:
When I look at this:

I'm thinking they want two equations. One for ##M## and one for ##m_1##.
No, that clause is to cover the fact that if the amplitude exceeds a bound then the objects will separate. This becomes relevant in a later part of the question.

Mister T
Update if anyone runs into the same problem. I don't have a solution but the attempt here is wrong. Right attempt would be ##x''m = -kx + m*g## Which leads to an inhom. diff. equation.

With this I close this thread o7

RiotRick said:
Update if anyone runs into the same problem. I don't have a solution but the attempt here is wrong. Right attempt would be ##x''m = -kx + m*g## Which leads to an inhom. diff. equation.

With this I close this thread o7
No, as I wrote in post #4 it depends how you define the position x=0. If you define it as being the equilibrium position then that is mg/k below the relaxed spring position. Thus the force in the spring is -k(x-mg/k) = -kx+mg. The net force on the object is thus (-kx+mg)-mg = -kx.