A mass m1 hangs from a spring k and is in static equilibrium. A second mass m2 drops through a height h and sticks to m1 without rebound. Determine the subsequent motion u(t) measured from the static equilibrium position of m1 and k.

Explanation:

From the information given:

The equation of the motion can be represented as:

where:

= mass of the body 1

= mass of the body 2

= acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency

And the equation for the general solution can be represented  as:

To determine the initial velocity, we have:

where h = height

Suppose we differentiate equation (2) with respect to time t; we have the following illustration:

now if t = 0

Then

Using the  law of conservation of momentum on the impact;

By replacing the value of with

Then the above equation becomes:

Making u(0) the subject of the formula, we have:

Similarly, the value of the variable can be determined as follows;

Using boundary conditions

Also, if  

Then :

making B the subject; we have:

Finally, replacing the value of A and B back to the general solution at equation (2); we have the equation of the subsequent motion u(t) which is:

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