Max Elongation of Mass-Spring System

[Moara]

Homework Statement

Find the maximum deformation of the spring in the system of the figure. F1 and F2 are Constant forces. Assume there is no friction and that the spring has no deformation in the begining of the motion

Relevant Equations

Fr=ma , Fel=kx, dW=Fdx

For the maximum elongation the velocities of the blocks should be equal. With that i used two conservation of energy for m1 and m2, for m1 the work of the elastic force is integral of kx(dx1) and for m2 kx(dx2), where X1+x2=x. I got three equation and four variables since i don't know the final speed, couldn't get any further

 

[Gaussian97]

I think conservation of energy is not necessary here, you can solve the problem only using Newton's second law.

 

[BvU]

Hi,

Assume there is no friction and that the spring has no deformation in the beginning of the motion

I suspect some more information is needed -- either that or I have to read 'in the beginning of the motion' as 'initially both blocks are at rest'.

For the maximum elongation the velocities of the blocks should be equal

and probably zero - so all the mechanical energy is in the spring... [edit2] not right. forget it.

You can easily exclude answer b by taking F1 = F2 and m1 = m2, which makes x = 0 and that is clearly not so.

Then: it is not very logical to use energy conservation if there are external forces at work. You only look at the stretching of the string and not at the work the forces do.

[edit] drat, Gauβ was a lot faster...

 

[Moara]

Sorry about the text, what i mean is they are at rest tio at the inicial moment. But i don't see why the final velocity should be zero. Furthermore, i tried using Newtons second law, but the integral of kxdx1, where x1 is the displacement of block m1 was difficult to execute

 

[Gaussian97]

Sorry about the text, what i mean is they are at rest tio at the inicial moment. But i don't see why the final velocity should be zero. Furthermore, i tried using Newtons second law, but the integral of kxdx1, where x1 is the displacement of block m1 was difficult to execute

I didn't make any integral in my calculus... Also, I don't think that the final velocity has to be 0.

 

[Moara]

did you use the reference frame of Earth?

 

[Gaussian97]

did you use the reference frame of Earth?

I think that, if we don't include relativity here, the result is RF independent.

 

[collinsmark]

[I temporarily deleted my post. I want to double check a few things first.]

 

[Moara]

Now i see, could finally do it , thanks

 

[Gaussian97]

Nice, what answer did you get?

 

[collinsmark]

Here is my previous post that I deleted to double check a few things. I've added a few comments to it. There's a certain part of this approach that's subtly tricky.
--------------------------------

I agree with @Gaussian97 that Newton's second law is the [might be a] better approach. (Although I did use integral calculus.)

Newton's second law states that the sum of all forces on a given mass equals that mass times its acceleration.
$$\sum_i \vec F_i = m \vec a$$

Set this up separately, one equation for $m_1$ and another for $m_2$

A couple things to keep in mind:

• The magnitude of the spring force on a given mass isn't the total displacement of the mass times the spring constant (e.g. $kx_1$), rather it is the total spring extension times the spring constant, $kx$. There's no need to split $x$ into $x_1$ and $x_2$ here. Just leave it as $x$. [Edit: You will need separate accelerations though, $a_1$ and $a_2$.]
• When you set up the equations, pay careful attention to the direction of each force. Pick a direction to represent the positive direction and stick with it. Consistency is important here.

From there, integrate to find the velocities $v_1$ and $v_2$.

[Edit: and here is the tricky bit. The spring's extension $x$ is not constant. It's a function of time. So if you integrate, say, $\int (F_1 - kx)dt$, the answer isn't simply $(F_1 - kx)t$. But the beauty here is that you don't have to actually evaluate the integrals. After setting the velocities equal to each other, differentiate both sides of the equation, thus getting rid of the integrals and the need to perform some nasty integration.]

And with your insight in your original post about the velocities being equal, simple algebra is all that's left.

 

[Gaussian97]

Here is my previous post that I deleted to double check a few things. I've added a few comments to it. There's a certain part of this approach that's subtly tricky.
--------------------------------

I agree with @Gaussian97 that Newton's second law is the [might be a] better approach. (Although I did use integral calculus.)

Newton's second law states that the sum of all forces on a given mass equals that mass times its acceleration.
$$\sum_i \vec F_i = m \vec a$$

Set this up separately, one equation for $m_1$ and another for $m_2$

A couple things to keep in mind:

• The magnitude of the spring force on a given mass isn't the total displacement of the mass times the spring constant (e.g. $kx_1$), rather it is the total spring extension times the spring constant, $kx$. There's no need to split $x$ into $x_1$ and $x_2$ here. Just leave it as $x$. [Edit: You will need separate accelerations though, $a_1$ and $a_2$.]
• When you set up the equations, pay careful attention to the direction of each force. Pick a direction to represent the positive direction and stick with it. Consistency is important here.

From there, integrate to find the velocities $v_1$ and $v_2$.

[Edit: and here is the tricky bit. The spring's extension $x$ is not constant. It's a function of time. So if you integrate, say, $\int (F_1 - kx)dt$, the answer isn't simply $(F_1 - kx)t$. But the beauty here is that you don't have to actually evaluate the integrals. After setting the velocities equal to each other, differentiate both sides of the equation, thus getting rid of the integrals and the need to perform some nasty integration.]

And with your insight in your original post about the velocities being equal, simple algebra is all that's left.

I think that the problem is much easier, I've solved without any integration.

 

[collinsmark]

I think that the problem is much easier, I've solved without any integration.

I didn't have to evaluate any integrals either in this problem. However, when using this Newton's second law approach, you do have to set up the integrals, at least as an interim step, to convert acceleration to velocity. $v = \int a \ dt$.

It's not sufficient to use a kinematics equation such as $v = at$ here since that only applies to uniform acceleration, and that doesn't apply here (the masses are oscillating).

 

[Gaussian97]

I didn't use any kinematics here, only dynamics and very little knowledge of the motion of a spring.