Two blocks and a spring system - Finding equation of motion

[Saitama]

Homework Statement

A system is composed of two blocks of mass $m_1$ and $m_2$ connected by a massless spring with spring constant k. The blocks slide on a frictionless plane. The unstretched length of the spring is $l$. Initially $m_2$ is held so that the spring is compressed to $l/2$ and $m_1$ is forced against a stop, as shown. $m_2$ is released at $t=0$.

Find the motion of the center of mass of the system as a function of time.

Homework Equations

The Attempt at a Solution

At any time t, let the distance of block $m_1$ from the wall be $x_1$ and that of $m_2$ be $x_2$. The extension in the spring length is $x_2-x_1-l$.
Applying Newton's second law on $m_1$,
$$k(x_2-x_1-l)=m_1\ddot{x_1} \Rightarrow \ddot{x_1}=\frac{k(x_2-x_1-l)}{m_1} (*)$$
Similarly,
$$\ddot{x_2}=-\frac{k(x_2-x_1-l)}{m_2} (**)$$
Subtracting (*) from (**) and substituting $z=x_2-x_1-l$ and $\mu=m_1m_2/(m_1+m_2)$,
$$\ddot{z}=-\frac{k}{\mu}z$$
Solution of the above differential equation is of the form,
$$z(t)=A\sin(\omega t)+B\cos(\omega t)$$
where $\omega=k/\mu$.
At t=0, z(0)=-l/2, z'(0)=0 hence, $A=0$ and $B=-l/2$. The equation of motion is
$$z(t)=-\frac{l}{2}\cos(\omega t)$$
Is the above equation correct? The question asks about the motion of center of mass and I am in a dilemma if the equation I have reached represents that. If not, what does the above equation represent then?

Any help is appreciated. Thanks!

 

[TSny]

Block 1 does not move for a while after the system is released - it remains pushed against the wall by the spring until the spring extends enough to pull the block to the right.

Since you are only interested in the motion of the center of mass, you can try using the famous theorem that the center of mass of a system moves like a point mass having a mass equal to the total mass of the system and acted on by a single force equal to the net external force acting on the system.

 

[Saitama]

Block 1 does not move for a while after the system is released - it remains pushed against the wall by the spring until the spring extends enough to pull the block to the right.

Since you are only interested in the motion of the center of mass, you can try using the famous theorem that the center of mass of a system moves like a point mass having a mass equal to the total mass of the system and acted on by a single force equal to the net external force acting on the system.

The net force acting on the system is the normal force on block $m_1$. The normal force varies as $m_2$ moves, I am unsure how to write it in equations.

Let the distance of $m_2$ be $x$ from the wall. Then change in length of spring is $l-x$. The normal force on $m_1$ is $k(l-x)$ to the right. Hence,
$$k(l-x)=(m_1+m_2)\ddot{x}_{CM}$$
$x_{CM}=m_2x/(m_1+m_2) \Rightarrow \ddot{x}_{CM}=m_2\ddot{x}/(m_1+m_2)$
$$\Rightarrow k(l-x)=m_2\ddot{x}$$.
This doesn't look right to me.

 

[TSny]

The net force acting on the system is the normal force on block $m_1$.

That's right.

The normal force varies as $m_2$ moves, I am unsure how to write it in equations.

If the left end of the spring were just attached directly to the wall, could you describe the motion of m₂ after it is released? With the left end attached to m₁ against the wall, would the motion of m₂ be any different up until the time m₁ leaves the wall?

Does that help to see how the spring force acting on m₁ will change with time? Will that help in finding the time dependence of the normal force acting on m₁?

 

[Saitama]

If the left end of the spring were just attached directly to the wall, could you describe the motion of m₂ after it is released?

Yes, it will be SHM.

With the left end attached to m₁ against the wall, would the motion of m₂ be any different up until the time m₁ leaves the wall?

No.

Does that help to see how the spring force acting on m₁ will change with time? Will that help in finding the time dependence of the normal force acting on m₁?

Won't the normal force be equal to the spring force for the time $m_1$ is attached to wall?

 

[TSny]

Won't the normal force be equal to the spring force for the time $m_1$ is attached to wall?

Yes, that's correct.

 

[Saitama]

Yes, that's correct.

Can you please check my equations in post #3?

 

[TSny]

Can you please check my equations in post #3?

Looks like the equations are correct to me. But I don't really like using $x$ for the position of mass 2 relative to the wall. Would you mind using $x_2$ instead? That way, you can let $x$ represent the amount of stretch of the spring from it's natural length. Then you can just write Hooke's law in the usual form $F = -kx$. Note that $x_2 = l+x$.

The reason for harping on the notation is that you are going to be interested in the external force from the wall which, as you already said, is determined by the spring force which is most simply expressed in terms of $x$.

Also, you should be able to write down the time dependence of $x$ immediately from your knowledge of simple harmonic motion. (At least up to the time that m₁ leaves the wall.) No need to set up and solve a differential equation, unless you just want to.

 

[TSny]

Just realized that since you can write down the stretch of the spring $x(t)$ from knowledge of SHM, you can then easily get $x_2(t)$ and therefore $x_{CM}(t)$ up until m₁ leaves the wall. After that, the motion of the CM is simple.

I solved it originally by working with the external force from the wall. But that's not necessary. Sorry for not seeing that earlier.

 

[Saitama]

Just realized that since you can write down the stretch of the spring $x(t)$ from knowledge of SHM, you can then easily get $x_2(t)$ and therefore $x_{CM}(t)$ up until m₁ leaves the wall. After that, the motion of the CM is simple.

I solved it originally by working with the external force from the wall. But that's not necessary. Sorry for not seeing that earlier.

The position of $m_2$ from the wall for the time $m_1$ is attached to wall can be represented by
$$x_2(t)=l\left(1-\frac{\cos(\omega t)}{2}\right)$$
where $\omega=\sqrt{k/m_2}$.
Hence, the position of CM of system,
$$x_{CM}=\frac{m_2l}{m_1+m_2}\left(1-\frac{\cos(\omega t)}{2}\right)$$
I am confused as to how I would represent the motion of CM after $m_1$ detaches from the wall? Do I follow the same approach as I did in my post #1? But that would give me the extension of spring as a function of time.

 

[voko]

I am confused as to how I would represent the motion of CM after $m_1$ detaches from the wall?

What happens with the CM when there are no external forces?

 

[TSny]

... $\omega=\sqrt{k/m_2}$.
Hence, the position of CM of system,
$$x_{CM}=\frac{m_2l}{m_1+m_2}\left(1-\frac{\cos(\omega t)}{2}\right)$$

That looks correct.

I am confused as to how I would represent the motion of CM after $m_1$ detaches from the wall? Do I follow the same approach as I did in my post #1? But that would give me the extension of spring as a function of time.

This is where the theorem about the motion of the center of mass that I cited earlier will help.

 

[Saitama]

The net external force on the system is zero. This means that the velocity of CM stays constant.

When $m_1$ is about to move, the velocity of $m_2$ is $l\omega/2$. The velocity of CM at that instant is $\displaystyle \frac{m_2l\omega}{2(m_1+m_2)}$ and this stays constant for the rest of the motion. Hence, the position of CM can be represented by
$$x_{CM}=\frac{m_2l\omega}{2(m_1+m_2)}\left(t+\frac{\pi}{2\omega}\right)$$

Correct?

EDIT: I think its better to write
$$x_{CM}=\frac{m_2l\omega}{2(m_1+m_2)}t$$

for $t \geq \pi/(2\omega)$.

 

[TSny]

EDIT: I think its better to write
$$x_{CM}=\frac{m_2l\omega}{2(m_1+m_2)}t$$

for $t \geq \pi/(2\omega)$.

That's now quite correct if you are measuring the position of the CM from the wall. Look at where that puts the CM at time $t = \pi/(2\omega)$

 

[Saitama]

That's now quite correct if you are measuring the position of the CM from the wall. Look at where that puts the CM at time $t = \pi/(2\omega)$

At $t=\pi/(2\omega)$,
$$x_{CM}=\frac{ml\pi}{4(m_1+m_2)}$$
This is wrong.

Does this mean the above applies only for $t>\pi/(2\omega)$?

 

[TSny]

What is the length of the spring when m₁ leaves the wall? Where does that put the CM at that instant?

 

[Saitama]

What is the length of the spring when m₁ leaves the wall?

It is unstretched when $m_1$ leaves the wall.

Where does that put the CM at that instant?

At a distance $m_2l/(m_1+m_2)$ from the wall.

 

[TSny]

At a distance $m_2l/(m_1+m_2)$ from the wall.

Right. But that doesn't agree with what you wrote for the location of the CM at $t = \pi/(2\omega)$.

 

[Saitama]

But that doesn't agree with what you wrote for the location of the CM at $t = \pi/(2\omega)$.

Yes. I modify my answer.
$$x_{CM}(t)=\frac{m_2l}{(m_1+m_2)}\left(1-\frac{\cos(\omega t)}{2}\right) \, \, \text{for} \, \, t\leq \frac{\pi}{2\omega}$$
$$x_{CM}(t)=\frac{m_2l\omega}{2(m_1+m_2)}t\, \, \text{for}\, \, t>\frac{\pi}{2\omega}$$
Is this correct?

 

[TSny]

Note that at $t =\frac{\pi}{2\omega}$ your solution for $t \leq \frac{\pi}{2\omega}$ does not agree with your solution for $t > \frac{\pi}{2\omega}$. Decide which one is correct, and then modify the other.

 

[Saitama]

Note that at $t =\frac{\pi}{2\omega}$ your solution for $t \leq \frac{\pi}{2\omega}$ does not agree with your solution for $t > \frac{\pi}{2\omega}$. Decide which one is correct, and then modify the other.

I still don't see where did I go wrong? Did I analyse the situation incorrectly in post #13?

 

[TSny]

In post #13, you got the correct expression for v_CM. The mistake is going from v_CM to x_CM.

Edit: Oh, I just noticed in post #14 that I typed the word "now" where I meant to type "not". Big difference . Hope that didn't throw you off.

 

[Saitama]

In post #13, you got the correct expression for v_CM. The mistake is going from v_CM to x_CM.

I tried different expressions so that at $t=\pi/(2\omega)$, both the expressions for $x_{CM}$ give the same value.
I came up with this:
$$x_{CM}(t)=\frac{m_2l\omega}{2(m_1+m_2)}\left(t-\frac{\pi-4}{2\omega}\right)$$
It gives the correct position for $t=\pi/(2\omega)$ but I reached this expression with hit and trial method which I feel is not the correct way.

Edit: Oh, I just noticed in post #14 that I typed the word "now" where I meant to type "not". Big difference . Hope that didn't throw you off.

Ah but that's my mistake. As a student, I should have asked if its "now" or "not". :)

 

[TSny]

I now (!) agree with your answer.

 

[Saitama]

I now (!) agree with your answer.

Umm..as I already said, I reached the expression with hit and trial, is there a better way?

 

[TSny]

If you do an indefinite integral with respect to time of the expression $v_{CM}$ for $t \geq \frac{\pi}{2\omega}$ you get $x_{CM}$ for $t \geq \frac{\pi}{2\omega}$ up to an arbitrary constant. Choose the constant so that $x_{CM}$ at $t = \frac{\pi}{2\omega}$ matches the solution for $t \leq \frac{\pi}{2\omega}$.