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?

 

[voko]

What is $z'(0)$ given the statement of the problem?

 

[Saitama]

What is $z'(0)$ given the statement of the problem?

$z=x_2-x_1-l_0 \Rightarrow z'(0)=x_2'(0)-x_1'(0)$, $x_1'(0)=0$ but I don't know about $x_2'(0)$, the force starts acting on $m_2$ at t=0 so $x_2'(0)$ should be also zero?

 

[voko]

You are sure that $m_1$ is stationary initially, but not so sure about $m_2$. Why?

 

[Saitama]

You are sure that $m_1$ is stationary initially, but not so sure about $m_2$. Why?

Sorry, I think I replied too hastily.

What should I do now? Is it possible to solve the D.E in such a way that it involves $z'$ too as it would be easy to solve then?

 

[voko]

I am not sure what you are trying to achieve now. You already have a general solution. All you need to do is use the initial conditions to fix two constants.

 

[Saitama]

I am not sure what you are trying to achieve now. You already have a general solution. All you need to do is use the initial conditions to fix two constants.

But how do I determine the constants? I do not know about $x_2'(0)$ and $x_1'(0)$.

 

[voko]

The statement of the problem has enough information on the velocities of both masses at t = 0. Just read it carefully.

 

[consciousness]

This calculus based approach is interesting but this question is easily solved by relative motion.

 

[voko]

This calculus based approach is interesting but this question is easily solved by relative motion.

Mea culpa, I really did think that solving a linear ODE was easy :)

In fact, we did get a relative motion equation (for z), it can be easily recast into an energy conservation law, from which the extrema can easily be deduced.

But at this stage I would leave that till after we are through the current solution.

 

[Saitama]

The statement of the problem has enough information on the velocities of both masses at t = 0. Just read it carefully.

The velocities are zero at t=0, i.e $z'(0)=0$.
At t=0, z=0, hence,
0=\frac{F\mu}{km_2}+B \Rightarrow B=-\frac{F\mu}{km_2}
Since z'(0)=0, A=0. Therefore,
z=\frac{F\mu}{km_2}(1-\cos(\omega t))
For maximum and minimum elongation, z'=0
z'=\frac{F\mu}{km_2}\sin(\omega t)=0
Hence, $\omega t=0,\pi$
Corresponding to these two values,
z=0,\frac{2F\mu}{km_2}

Thanks for the help and patience voko!

This calculus based approach is interesting but this question is easily solved by relative motion.

How? A few hints would be great. :)

 

[consciousness]

Analyse the FBD of m2 w.r.t. m1.

The forces acting on it are-
1)F towards right.
2)kx where x is the extension in the spring towards left.
3)A pseudo force towards left as m1 is accelerating.

I think it was an SHM.

 

[sankalpmittal]

Analyse the FBD of m2 w.r.t. m1.

The forces acting on it are-
1)F towards right.
2)kx where x is the extension in the spring towards left.
3)A pseudo force towards left as m1 is accelerating.

I think it was an SHM.

Good. Applying free body diagram method from the non inertial frame of block is really nice. Take m₂ to be the system and assume non inertial frame to be attached at it. This will simplify the problem to some extent.

 

[Saitama]

Analyse the FBD of m2 w.r.t. m1.

The forces acting on it are-
1)F towards right.
2)kx where x is the extension in the spring towards left.
3)A pseudo force towards left as m1 is accelerating.

I think it was an SHM.

Applying Newton's Second law on $m_2$,
$$F-kx-m_1a_1=m_2a_2$$
How should I solve this (without calculus)? Is it possible to do it without the relative motion? I mean only by using energy conservation?
$$F(x_2-l_0)=\frac{1}{2}k(x_2-x_1-l_0)^2$$
I need more equations here.

 

[ehild]

m1 also performs simple harmonic motion. No sense to use a frame of reference attached to it.

You can describe the motion with respect to the CM. You know that the CM moves under the effect of the external force(s). It is F, a single constant force now, so the CM moves with uniform acceleration a_cm=F/(m1+m2). That means pseudo forces -m₁a_cm and -m₂a_cm acting on the masses. You need two equations again in order to describe the accelerations of the bodies.Dividing by the masses and subtracting the equations and introducing the variable z as before, you get the same differential equation as before.
The solution must be familiar to you: it is the same as that of a mass hanging on a spring. A constant force, added to the spring force. Instead of the hanging mass, you have μ=m1m2/(m1+m2) , the reduced mass of the two bodies. The solution is an SHM about a constant z.

ehild

 

[Saitama]

m1 also performs simple harmonic motion. No sense to use a frame of reference attached to it.

You can describe the motion with respect to the CM. You know that the CM moves under the effect of the external force(s). It is F, a single constant force now, so the CM moves with uniform acceleration a_cm=F/(m1+m2). That means pseudo forces -m₁a_cm and -m₂a_cm acting on the masses. You need two equations again in order to describe the accelerations of the bodies.Dividing by the masses and subtracting the equations and introducing the variable z as before, you get the same differential equation as before.
The solution must be familiar to you: it is the same as that of a mass hanging on a spring. A constant force, added to the spring force. Instead of the hanging mass, you have μ=m1m2/(m1+m2) , the reduced mass of the two bodies. The solution is an SHM about a constant z.

ehild

Thanks ehild but why can't I use conservation of energy here?