Force on a two block and a spring system

[ehild]

I did not say that you can't use it. Try. I am eager to see how the problem can be solved with conservation of energy in a simple way. I have no idea. Teach me.

ehiléd

 

[Saitama]

Teach me.

At maximum and minimum elongation, kinetic energy of system is zero. Working in the frame of reference fixed to CM, let $x_2$ be the distance moved by $m_2$ towards right and $x_1$ be the distance moved by $m_1$ towards [strike]right[/strike] left. The force acting on $m_2$ are F (towards right) and $m_2a_{cm}$ (towards left). The net force on $m_2$ is $m_1F/(m_1+m_2)$. The net force acting on $m_1$ is $m_1a_{cm}=m_1F/(m_1+m_2)$ (towards left).
From conservation of energy,
\frac{m_1F}{m_1+m_2}x_1+\frac{m_1F}{m_1+m_2}x_2=\frac{1}{2}k(x_1+x_2)^2
Solving for $x_1+x_2$, I get two values,
x_1+x_2=0, \frac{2m_1F}{k(m_1+m_2)}

But is it possible to do this in the reference frame fixed to the ground?

 

[voko]

The only other frame that makes sense to me is that of the center of mass. I do not think that makes the problem much simpler, though.

 

[ehild]

At maximum and minimum elongation, kinetic energy of system is zero.

You mean in the CM frame of reference. As the whole system moves with constant acceleration, never in rest...

Working in the frame of reference fixed to CM, let $x_2$ be the distance moved by $m_2$ towards right and $x_1$ be the distance moved by $m_1$ towards right. The force acting on $m_2$ are F (towards right) and $m_2a_{cm}$ (towards left).
From conservation of energy,
\frac{m_1F}{m_1+m_2}x_1+\frac{m_1F}{m_1+m_2}x_2=\frac{1}{2}k(x_1+x_2)^2
Solving for $x_1+x_2$, I get two values,
x_1+x_2=0, \frac{2m_1F}{k(m_1+m_2)}

Clever. Congratulation!

But is it possible to do this in the reference frame fixed to the ground?

I do not have the slightest idea. I prefer solving differential equations. But at maximum and minimum distances, the velocity of the masses is the same as that of the CM.

ehild

 

[voko]

From conservation of energy

I think you have sign errors in what follows. You are looking for $x_1 - x_2$, not for $x_1 + x_2$.

 

[consciousness]

Applying Newton's Second law on $m_2$,
$$F-kx-m_1a_1=m_2a_2$$

As we are analyzing wrt m1 it should be
$$F-kx-m_2a_1=m_2a_2$$
To find a1-

$$kx=m_1a_1$$
$$a_1=kx/m_1$$

then we get

$$Fnet=F-kx(1+m_2/m_1)$$

Now we can find the mean position(when net force is zero) of the SHM and its amplitude. As we are analyzing wrt m1 it is at rest for us. The maximum elongation of the spring is when m2 is as far away from m1 as possible.

 

[Saitama]

I think you have sign errors in what follows. You are looking for $x_1 - x_2$, not for $x_1 + x_2$.

No, I was calculating $x_1+x_2$. I explained in my previous post what $x_2$ and $x_1$ are.

 

[voko]

No, I was calculating $x_1+x_2$. I explained in my previous post what $x_2$ and $x_1$ are.

You defined both variables as distances toward right (from the center of mass, apparently). You are looking for the distance between the masses. That cannot be $x_1 + x_2$ with your definition.

 

[Saitama]

You defined both variables as distances toward right (from the center of mass, apparently). You are looking for the distance between the masses. That cannot be $x_1 + x_2$ with your definition.

Woops, sorry, I meant $x_1$ to the left.