Question about two falling masses connected with a spring

[Mohammad]

Hi, this is my first post :)

I have a question in reference to the same problem in: https://www.physicsforums.com/showthread.php?t=67697&highlight=Falling+Mass

I am having trouble comprehending the 3rd part.

I understand that the differential equation to solve is:
$$m\frac{d^2y}{dt} + 2ky = mg$$

Solving this equation yields the following complete solution:
$$z(t) = \frac{mg}{2k} + c_1cos(\sqrt{2k/m}t) + c_2sin(\sqrt{2k/m}t)$$

I am stuck at this point. The solution the author posted in the above-mentioned topic doesn't seem coherent to me from a mathematical point of view. Any guidance would be appretiated.

 

[learningphysics]

Hi, this is my first post :)

I have a question in reference to the same problem in: https://www.physicsforums.com/showthread.php?t=67697&highlight=Falling+Mass

I am having trouble comprehending the 3rd part.

I understand that the differential equation to solve is:
$$m\frac{d^2y}{dt} + 2ky = mg$$

It appears to me that the differential equation should be:
$$m\frac{d^2z}{dt} + 2kz = 0$$

Where z is the distance of the bottom mass from the center of mass of the system.

The acceleration of the bottom mass is (mg-kx)/m and the acceleration of the entire system is g. So the relative acceleration is (mg-kx)/m - g
=-kx/m = -2kz/m (since z=1/2 the total extension x). a=-2kz/m leads to the diff. eq above.

Complete solution is (as you posted but without one term):
$$z(t)=c_1cos(\sqrt{2k/m}t) + c_2sin(\sqrt{2k/m}t)$$

Solving this with z(0)=mg/2k and z'(0)=0 gives the particular solution:

$$z(t)=(mg/2k) cos(\sqrt{2k/m}t)$$

 

[Mohammad]

Thank you for your reply. I can understand what is going on now. I was basically mixing up the acceleration relative to the ground with the acceleration relative to the center of mass.