Hanging Spring with non-negligible mass is subjected to a driving force

[LCSphysicist]

Homework Statement

Print below

Relevant Equations

I will post below.

First i will use a equation resulted in considering a spring as a continuum limit of massive mass:

ω = √(KL/ρ)*kn

ρ is the linear density ρ

ω = √(KL/ρ)*kn

X = A*cos(kn*x)*cos(ω*t)

ξ = A (first consequence)

X = ξ*cos(kn*x)*cos(ω*t)

∂y/∂x need to be zero in x=L (for strain be zero)

kn*L = n*π (second)

ω = √(KL/ρ)*kn = √(KL/ρ)*n*π/L

So

X = L + ξ*cos(((n*π))*cos(ω*t)

where

ω = √(K/M)*n*π

n is a natural

This seems okay?

 

[haruspex]

Homework Statement:: Print below
Relevant Equations:: I will post below.

First i will use a equation resulted in considering a spring as a continuum limit of massive mass:

You've lost me already. I don't know what you mean by that, nor how you get that equation, nor what k_n is. For the equation to be dimensionally consistent, k_n would have to have dimension L^-1, right?

I would expect to see a differential equation obtained by considering an element length dy subject to different tensions above and below.

 

[LCSphysicist]

You've lost me already. I don't know what you mean by that, nor how you get that equation, nor what k_n is.
I would expect to see a differential equation obtained by considering an element length dy subject to different tensions above and below.

I am trying to approach the problem by means of waves, so when the up part is oscillating, the wave will propagate by the spring until the end of the spring...
I think i need to explain better, kn has the same sense as we use it in string oscillating vertically, it is like the wave number.
And the expression for w is just a result by oscillate the massive spring:

Imagine a spring as a lot of masses in line and as perturbation in one end is made.

You know the result ω will be

But as we are imagining a massive spring,

m is the little mass
Ka is the spring stiffness in the little mass region.

a -> 0

When we cut a spring, the stiffnes of each part is 2K right? Following this thinking:

Ka*a = K*L

ω² = (4*K*L/m*a)*k²*a²/4

ω = √(K*L/ρ)*k
ρ = M/L

and just put in the stationary wave with respectives conditions...

Since the waves propagates in the spring direction, when it came below, it will governs the motion of the end

 

[haruspex]

Ok, I understand a little better, but it doesn’t seem to me you have arrived at an answer to the question. I would have thought we are supposed to find the phase and amplitude at the free end.
Also, I am confused by your use of x. In one place it looks like x is the distance of an element from one end in the relaxed state, and y is its displacement from its equilibrium position; in another, x seems to be the position of an element as a function of time.
Please define all your variables clearly.

 

[LCSphysicist]

Ok, I understand a little better, but it doesn’t seem to me you have arrived at an answer to the question. I would have thought we are supposed to find the phase and amplitude at the free end.
Also, I am confused by your use of x. In one place it looks like x is the distance of an element from one end in the relaxed state, and y is its displacement from its equilibrium position; in another, x seems to be the position of an element as a function of time.
Please define all your variables clearly.

I agree there is a flaw in approach the problem by this way, because i can not find any k, and i don't know how could we find it.

I will rewrite:

Now y denote the initial position and X the position at time t

ω = √(KL/ρ)*kn
ρ is the linear density ρ

X = y + A*cos(kn*y)*cos(ω*t)

(ξ*cos(ω*t) at y=0)=>

X = ξ*cos(kn*y)*cos(ω*t)

∂X/∂y need to be zero in y=L (for strain be zero)

kn*L = n*π (second)

ω = √(KL/ρ)*kn = √(KL/ρ)*n*π/L

So

X = L + ξ*cos(((n*π))*cos(ω*t)

where
ω = √(K/M)*n*π

 

[Chestermiller]

If the spring were deformed uniformly, the tension would be $$T=Ku_L$$where $u_L$ is the displacement at x = L, assuming that x = 0 is fixed. This equation can also be written as $$T=KL\frac{u_L}{L}$$where the local uniform strain in the spring is given by $$\epsilon(x)=\frac{u_L}{L}=\frac{\partial u}{\partial x}$$Therefore, in any arbitration non-uniform axial deformation of the spring, the local tension is given by:
$$T(x)=KL\frac{\partial u}{\partial x}$$

For the problem at hand, if we perform a force balance on the portion of the spring between x and $x+\Delta x$, we obtain:$$T(x+\Delta x)-T(x)=\frac{m}{L}\Delta x\frac{\partial^2u}{\partial t^2}$$Combining the previous two equations, dividing by $\Delta x$, and taking the limit as $\Delta x$ approaches zero yields:
$$\frac{\partial^2u}{\partial t^2}=\frac{KL^2}{m}\frac{\partial^2u}{\partial x^2}=c^2\frac{\partial^2u}{\partial x^2}$$where c is the wave speed: $$c=L\sqrt{\frac{K}{m}}$$The boundary condition at x = L is zero tension: $\frac{\partial u}{\partial x}=0$ @ x = L.

The solution to this problem is going to be of the form $$u(x,t)=A(x)\cos{\omega t}+B(x)\sin{\omega t}$$subject to the initial conditions $A(0)=\epsilon$, B(0)=0.

 

[etotheipi]

@Chestermiller just to clarify, have you defined $x$ to be the material coordinates and $u(x,t)$ as the displacement of the point of the spring at material coordinate $x$ from its position at $t=0$? That is to say that the function which converts the material coordinate to the 'real' coordinate $X$ is $X(x,t) = x + u(x,t)$?

 

[haruspex]

The solution to this problem is going to be of the form $$u(x,t)=A(x)\cos{\omega t}+B(x)\sin{\omega t}$$

I worried that there might be harmonics. Is that not possible ?

 

[LCSphysicist]

If the spring were deformed uniformly, the tension would be $$T=Ku_L$$where $u_L$ is the displacement at x = L, assuming that x = 0 is fixed. This equation can also be written as $$T=KL\frac{u_L}{L}$$where the local uniform strain in the spring is given by $$\epsilon(x)=\frac{u_L}{L}=\frac{\partial u}{\partial x}$$Therefore, in any arbitration non-uniform axial deformation of the spring, the local tension is given by:
$$T(x)=KL\frac{\partial u}{\partial x}$$

For the problem at hand, if we perform a force balance on the portion of the spring between x and $x+\Delta x$, we obtain:$$T(x+\Delta x)-T(x)=\frac{m}{L}\Delta x\frac{\partial^2u}{\partial t^2}$$Combining the previous two equations, dividing by $\Delta x$, and taking the limit as $\Delta x$ approaches zero yields:
$$\frac{\partial^2u}{\partial t^2}=\frac{KL^2}{m}\frac{\partial^2u}{\partial x^2}=c^2\frac{\partial^2u}{\partial x^2}$$where c is the wave speed: $$c=L\sqrt{\frac{K}{m}}$$The boundary condition at x = L is zero tension: $\frac{\partial u}{\partial x}=0$ @ x = L.

The solution to this problem is going to be of the form $$u(x,t)=A(x)\cos{\omega t}+B(x)\sin{\omega t}$$subject to the initial conditions $A(0)=\epsilon$, B(0)=0.

I almost get it, unless the X = L + ξ*cos(((n*π))*cos(ω*t) in the equation, so in my way n would be some even number? Why should it be?

 

[LCSphysicist]

This remember the fact when we driven something below resonance frequency, the system resonates with the force, while above the resonance frequency, it will oscillates180 degree out.

 

[Chestermiller]

@Chestermiller just to clarify, have you defined $x$ to be the material coordinates and $u(x,t)$ as the displacement of the point of the spring at material coordinate $x$ from its position at $t=0$? That is to say that the function which converts the material coordinate to the 'real' coordinate $X$ is $X(x,t) = x + u(x,t)$?

Yes, material point, but not at time zero. It is supposed to be distance along the spring at a time when the spring is not under any load. In the present system, the motion is supposed to be a stationary oscillatory motion, so that at each material point, it is periodic, but with a different phase angle.

 

[Chestermiller]

I worried that there might be harmonics. Is that not possible ?

I don't think that higher harmonics can satisfy the boundary conditions.

 

[Chestermiller]

I almost get it, unless the X = L + ξ*cos(((n*π))*cos(ω*t) in the equation, so in my way n would be some even number? Why should it be?

I don't get this result. I get the following standing wave:
$$u(x,t)=\epsilon\frac{\cos{\frac{\omega}{c}(L-x)}}{\cos{\frac{\omega}{c}L}}\cos{\omega t}$$I believe that this satisfies the boundary conditions and the differential equation.