Mechanical Vibrations problem

[naggy]

Homework Statement

A mass m is attached to a spring(massless) that is located inside a massless box. The box is falling under gravity. When the box starts to fall the spring is in it's equilibrium position and the box sticks to the ground when it hits it.

-The box is a distance H from the ground
-Spring has spring constant k
-The mass on the spring is m



Find the equation of motion (and initial conditions) when
a)the box is falling and
b)when the box has landed.

Variables
x is movement from equilibrium position of spring
y is distance from ground to mass

Homework Equations

$$L=KE - PE$$
or
$$F=m\ddot{x}$$

The Attempt at a Solution

I prefer using Lagrangian equations. When the box is falling:
$$KE= \frac{1}{2}m\dot{x^2}$$
$$PE= mgy +\frac{1}{2}kx^2$$

Now can I connect y(distance from the ground to m) and x(movement from equilibrium position of mass) with y=constant + x and use the Euler lagrange equations?

I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion

when the box lands, maybe x(tH)=H and x'(tH)=sqrt(2gH) ??

 

[Imperitor]

Homework Statement

I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion

Just think what happens when the box hits the ground. It will stop but the mass on the spring will still have same velocity because nothing is stopping it. The only contribution of the fall on the system is an initial velocity. Hope that helps.

 

[Mene]

I would first clarify the problem by asking for more information. Is the box falling with a constant velocity or is it accelerating? Is there any damping in the system? What is the initial position and velocity of the mass and the box? Without this information, it is difficult to provide a complete and accurate response.

Assuming that the box is falling with a constant velocity and there is no damping, the equation of motion for this system can be derived using the Lagrangian method. The Lagrangian for this system is given by:

L = T - V

where T is the kinetic energy and V is the potential energy. The kinetic energy of the system is given by:

T = 1/2 * m * (dx/dt)^2

where m is the mass of the object attached to the spring and dx/dt is the velocity of the object. The potential energy of the system is given by:

V = mgy + 1/2 * kx^2

where g is the acceleration due to gravity and k is the spring constant.

Using the Euler-Lagrange equations, we can find the equation of motion for this system:

d/dt (dL/dx') - dL/dx = 0

Substituting the expressions for T and V, we get:

m * d^2x/dt^2 + kx = 0

This is a second-order differential equation, and the solution depends on the initial conditions. As mentioned in the attempt at a solution, the initial conditions for the box falling would be x(0) = 0 and dx/dt(0) = 0.

Once the box has landed, the equation of motion will change, as the mass will no longer be in free fall. The new equation of motion will be:

m * d^2x/dt^2 + kx = mg

where mg is the weight of the mass. The initial conditions for this scenario would be x(0) = H and dx/dt(0) = 0, as the mass will be at a distance H from the ground and will have zero initial velocity.

In conclusion, the equation of motion for this system can be found using the Lagrangian method. However, more information is needed to accurately solve the problem and determine the initial conditions.