Force on a two block and a spring system

[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?

 

[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.