Force on a two block and a spring system

[Saitama]

Homework Statement

Two bars connected by a weightless spring of stiffness $k$ and length (in the non-deformed state) $\ell_o$ rest on a horizontal plane. A constant horizontal force $F$ starts acting on one of the bars as shown in the figure. Find the maximum and minimum distance between the bars during the subsequent motion of the system, if the masses of the bars are:
(a)equal
(b)equal to $m_1$ and $m_2$ and the force $F$ is applied to the bar of mass $m_2$.

Answers:
a)$l_{max}=l_o+F/k$ and $l_{min}=l_o$
b)$l_{max}=l_0+2m_1F/(k(m_1+m_2))$ and $l_{min}=l_o$

Homework Equations

The Attempt at a Solution

Case a):
From energy conservation,
$$Fx=\frac{1}{2}kx^2+\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2$$
where $x$ is the extension of the spring, $m$ is the mass of the blocks, $v_1$ and $v_2$ are the velocities of the blocks.
I still need one more equation to relate $v_1$ and $v_2$.

Thanks!

 

[Tanya Sharma]

Case a):
From energy conservation,
$$Fx=\frac{1}{2}kx^2+\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2$$
where $x$ is the extension of the spring, $m$ is the mass of the blocks, $v_1$ and $v_2$ are the velocities of the blocks.

Why should you go for energy conservation ?

Think for a while,how would the two blocks be moving with respect to each other when separation between them is minimum and when it is maximum .If you are able to figure out this,the problem becomes easy.

 

[voko]

There could be a smarter way, but if everything else fails, just write down the diff. eq.'s of the system and solve them. Should not be too hard.

 

[Saitama]

There could be a smarter way, but if everything else fails, just write down the diff. eq.'s of the system and solve them. Should not be too hard.

How should I write those diff eq.'s, can I have a few hints? :)

 

[voko]

Start with a FBD for each mass.

 

[Saitama]

Start with a FBD for each mass.

I will start with part b) first.

Forces (except the weight and normal reaction from ground) acting on $m_2$ are $F$ and force due to spring i.e $kx$ where $x$ is the extension of spring.
$$F-kx=m_2a_2$$

The only force acting on $m_1$ is $kx$
$$kx=m_1a_1$$

From the two equations, $$F=m_1a_1+m_2a_2$$
Should I replace $a_1$ with $dv_1/dt$ and $a_2$ with $dv_2/dt$? Is the above equation correct?

 

[voko]

Yes, your equations are correct, assuming you can express $x$ via $x_1, \ x_2$ and $l_0$.

 

[Saitama]

Yes, your equations are correct, assuming you can express $x$ via $x_1, \ x_2$ and $l_0$.

$x=x_1+x_2-l_0$?

So $a_1=d^2x_1/dt$ and $a_2=d^2x_2/dt$?

 

[voko]

Let's say $x_1 = 0$ and $x_2 = l_0$. Then $x = 0$ as one would expect - the spring is relaxed. Now let's move $m_1$ toward $m_2$, say, by letting $x_1 = l_0/2$. Then $x = l_0/2$, i.e., greater than in the relaxed state, while it should be less.

 

[consciousness]

I would prefer to do this problem by analyzing the motion of m2 w.r.t. m1. This removes the hassle of two variables x1,x2 leaving only one whose maximum we have to find. Remember to make the FBD of m2 carefully though.

 

[Saitama]

Let's say $x_1 = 0$ and $x_2 = l_0$. Then $x = 0$ as one would expect - the spring is relaxed. Now let's move $m_1$ toward $m_2$, say, by letting $x_1 = l_0/2$. Then $x = l_0/2$, i.e., greater than in the relaxed state, while it should be less.

I thought more about it. The best I could come up with is $x=x_2-x_1-l_0$. Is this correct?

Also, where should I use it?

I have $F=m_1a_1+m_2a_2$ but I don't have $x$ here.

 

[voko]

Check if your equations make sense. Keep one variable fixed, then another one. Is that consistent with the simple "one mass" case? Pay attention to the signs.

 

[Saitama]

Check if your equations make sense. Keep one variable fixed, then another one. Is that consistent with the simple "one mass" case? Pay attention to the signs.

You mean to say my equation for x is incorrect? I can't think of anything else. :(

If I consider your cases (post #9), when $x_1=0$ and $x_2=l_0$, $x=0$ i.e for this case, my equation is correct. When $m_2$ is fixed and $m_1$ is moved a distance $l_0/2$ i.e $x_2=l_0$ and $x_1=l_0/2$, $x=-l_0/2$, this is also correct, I can't see where my equation is wrong.

 

[voko]

I am not saying anything is incorrect. I am saying that I would like you to check whether your equations in #6 are consistent with the your definition of x.

 

[Saitama]

I am not saying anything is incorrect. I am saying that I would like you to check whether your equations in #6 are consistent with the your definition of x.

But still, where should I use this equation for $x$?

 

[voko]

Did I not say #6?

 

[Saitama]

Did I not say #6?

The final equation I had was $F=m_1a_1+m_2a_2$ but there is no x in this.

 

[voko]

Does that make it automatically correct or even useful to begin with?

 

[Saitama]

Does that make it automatically correct or even useful to begin with?

I am still not sure what to do. I will begin with using $kx=m_1a_1$ as it involves $x$.
$$k(x_2-x_1-l_0)=m_1\frac{d^2x_1}{dt^2}$$

I can replace $x$ and $a_2$ in the equation, $F-kx=m_2a_2$ but what I am supposed to do with these equations?

 

[voko]

Hmm. I think I outlined the general approach in #3, no?

 

[Saitama]

Hmm. I think I outlined the general approach in #3, no?

Subtracting the two equations I had,
$$F-2kx=m_1a_1-m_2a_2$$
$$m_2x_2''-m_1x_1''=F-2k(x_2-x_1-l_0)$$

I don't know how to solve this. :(

 

[voko]

Not too useful, but you are thinking along the right lines. Let $z = x_2 - x_1$. See if you can combine the equations so that you end up with just one equation for $z$ only.

 

[Saitama]

Not too useful, but you are thinking along the right lines. Let $z = x_2 - x_1$. See if you can combine the equations so that you end up with just one equation for $z$ only.

I did think of something similar but instead, I used the substitution, $z=x_2-x_1-l_0$ so that I could replace $x_2''-x_1''=z''$ but I here have $m_2x_2''-m_1x_1''$ so my substitution fails here.

 

[voko]

You are not trying hard enough :)

Try $m_1 m_2 x_2'' - m_1 m_2 x_1''$.

 

[Saitama]

Try $m_1 m_2 x_2'' - m_1 m_2 x_1''$.

I still cannot figure this out. What you wrote is equal to $m_1m_2z''$ but I have $m_2x_2''-m_1x_1''$.

 

[voko]

Are you saying you cannot use the equations for $x_1$ and $x_2$ to figure out what the expression I gave in #24 equals to?

 

[Saitama]

Are you saying you cannot use the equations for $x_1$ and $x_2$ to figure out what the expression I gave in #24 equals to?

You asked me to write $x_2-x_1=z$, from this $x_2''-x_1''=z''$. So $m_2m_1x_2''-m_2m_1x_1''=m_1m_2(x_2''-x_1'')=m_1m_2z''$.

 

[voko]

You have written this a few times by now, but you never used the equations for $x_1$ and $x_2$, even though I hinted quite heavily. Why?

 

[Saitama]

You have written this a few times by now, but you never used the equations for $x_1$ and $x_2$, even though I hinted quite heavily. Why?

What equations for $x_1$ and $x_2$ are you talking about?

I had $x=x_2-x_1-l_0$, from here $x''=x_2''-x_1''$. Is this the equation?

 

[voko]

You have derived equations relating $x_1''$ and $x_2''$ with $x_1$ and $x_2$. These are the primary equations describing the system, but you keep ignoring them.

 

[Saitama]

You have derived equations relating $x_1''$ and $x_2''$ with $x_1$ and $x_2$. These are the primary equations describing the system, but you keep ignoring them.

I think writing the equations in a different way would help.
$$\frac{kx}{m_1}=x_1''$$
and $$\frac{F}{m_2}-\frac{kx}{m_2}=x_2''$$

Subtracting the equations,
$$\frac{F}{m_2}-k\left(\frac{1}{m_1}+\frac{1}{m_2}\right)(x_2-x_1-l_0)=x_2''-x_1''$$
Substituting $x_2-x_1-l_0=z$,
$$\frac{F}{m_2}-kz\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=z''$$

Looks correct now? :)

 

[voko]

Looking good, but why stop here?

 

[Saitama]

Looking good, but why stop here?

You know it, I don't know how to solve second order D.Es.

I plugged a similar equation in wolfram alpha, is the following correct?
http://www.wolframalpha.com/input/?i=a-by=y''

 

[voko]

If you don't know how to solve a diff. eq.,, you should at least be able to check whether a proposed solution is correct, I would think.

 

[Saitama]

If you don't know how to solve a diff. eq.,, you should at least be able to check whether a proposed solution is correct, I would think.

Assuming what wolfram alpha gave me is correct.
$$z=\frac{F \mu}{km_2}+A\sin(\omega t)+B\cos(\omega t)$$
where $\mu=m_1m_2/(m_1+m_2)$, $\omega=\sqrt{k/\mu}t$ and $A$ and $B$ are constants.

I can determine one of the constant from the condition that at t=0, z=0. How would I determine the second constant?