Answer:
[tex]\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}[/tex]
Explanation:
From the information given:
The equation of the motion can be represented as:
[tex](m_1 +m_2) \hat u + ku = m_2 g--- (1)[/tex]
where:
[tex]m_1[/tex] = mass of the body 1
[tex]m_2[/tex] = mass of the body 2
[tex]\hat u[/tex] = acceleration
k = spring constant
u = displacement
g = acceleration due to gravity
Recall that the formula for natural frequency [tex]\omega _n = \sqrt{\dfrac{k}{m_1+m_2}}[/tex]
And the equation for the general solution can be represented as:
[tex]u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)[/tex]
To determine the initial velocity, we have:
[tex]\hat u_2^2 = 2gh[/tex]
[tex]\hat u_2 = \sqrt{2gh}[/tex]
where h = height
Suppose we differentiate equation (2) with respect to time t; we have the following illustration:
[tex]\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0[/tex]
now if t = 0
Then
[tex]\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0[/tex]
[tex]= \omega _n B[/tex]
Using the law of conservation of momentum on the impact;
[tex]m_2 \hat u_2=(m_1+m_2) \hat u (0)[/tex]
By replacing the value of [tex]\hat u_2[/tex] with [tex]\sqrt{2gh[/tex]
Then the above equation becomes:
[tex]m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)[/tex]
Making u(0) the subject of the formula, we have:
[tex]u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}[/tex]
Similarly, the value of the variable can be determined as follows;
Using boundary conditions
[tex]0 = A cos 0 + B sin 0 + \dfrac{m_2g}{k}[/tex]
[tex]0 = A (1)+0+ \dfrac{m_2g}{k}[/tex]
[tex]A =- \dfrac{m_2g}{k}[/tex]
Also, if [tex]\hat u (0) = \omega_nB[/tex]
Then :
[tex]\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B[/tex]
making B the subject; we have:
[tex]B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}[/tex]
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:
[tex]\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}[/tex]
