Question

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.

Answer 1

Answer:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$

Explanation:

From the information given:

The equation of the motion can be represented as:

$(m_1 +m_2) \hat u + ku = m_2 g--- (1)$

where:

$m_1$ = mass of the body 1

$m_2$ = mass of the body 2

$\hat u$ = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency $\omega _n = \sqrt{\dfrac{k}{m_1+m_2}}$

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

$u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)$

To determine the initial velocity, we have:

$\hat u_2^2 = 2gh$

$\hat u_2 = \sqrt{2gh}$

where h = height

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

$\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0$

now if t = 0

Then

$\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0$

$= \omega _n B$

Using the  law of conservation of momentum on the impact;

$m_2 \hat u_2=(m_1+m_2) \hat u (0)$

By replacing the value of $\hat u_2$ with $\sqrt{2gh$

Then the above equation becomes:

$m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)$

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

$u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}$

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

Using boundary conditions

$0 = A cos 0 + B sin 0 + \dfrac{m_2g}{k}$

$0 = A (1)+0+ \dfrac{m_2g}{k}$

$A =- \dfrac{m_2g}{k}$

Also, if  $\hat u (0) = \omega_nB$

Then :

$\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B$

making B the subject; we have:

$B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}$

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:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$