Distribution of Energy when work is done on a system of 2 masses connected by a spring

[MattGeo]

[Mentors’ note: No template because this post was moved from the technical forums. Everything that the template asks for seems to be present in the body of the post]

Suppose there is a spring-mass system arranged as shown in my crude drawing. This occurs on a frictionless surface. The spring is 0.5 meters long and is at its natural length. The 2 masses are initially at rest and the left mass is 1 kg and the right mass is 3 kg. If a 10 N force is applied leftward as shown on the 3 kg mass, then what will be the kinetic energy of the system once the force has been applied through a displacement of 10 meters? Additionally, what will be the potential energy stored in the spring due to its compression, assuming spring constant is 0.5 N/m? What will be the spring force at maximum compression?

My first inclination was to calculate the spring force at maximum compression and the magnitude of its compression. I arrived at this calculating the acceleration of the center of mass of the system.

Applying Newton's second law to the entire system the block system should accelerate at 2.5 m/s^2 and then using this fact you can deduce that the net force on the 1 kg1 kg block is 2.5 N and the net force on the 3 kg block is 7.5 N, meaning that the spring force at maximum compression is 2.5 N.

By considering the average force exerted by the spring to be 1.25 N I concluded that there must be 6.25 J of energy stored in the spring and 93.75 J of energy showing up as kinetic energy of the 2-mass system after 10 m of displacement. I feel like this must be incorrect though. I can't tell if I am even thinking about the displacement correctly. The spring is being compressed while the system begins to move.

Application of the force to the 3 kg mass will cause it to begin to accelerate but the moment it does, it will start applying a force to the other block, causing it to accelerate, with some tiny lag before it starts moving of course.

As long as the 10 N force is continuously applied and the acceleration is constant, would the spring compress to a stable compressed length or would it actually undergo oscillations while being accelerated?

Upon further thought I think I should be considering the displacement of the center of mass but also the relative displacement between the 2 masses and assume that the center of mass moves in proportion to m2/(m1+m2)

I keep confusing myself with this problem but if anyone could help to walk me through it, it would be great.

 

[ergospherical]

The net force on the whole system ($F=-10\mathrm{N}$) governs the acceleration of the centre of mass, which is $a_{\mathrm{com}} = F/(m_1 + m_2)$ and a constant.

You should transform into the (accelerating) centre of mass frame. Although the length of the spring can vary with time, $l = l(t)$, the centre of mass of the system -- i.e. the point that is a fraction $m_2/(m_1 + m_2)$ along the length of the spring, can be considered fixed in this accelerating frame. So you now essentially have two independent springs on either side of this point.

Now you can analyze the accelerations of the two masses from within this accelerating frame, remembering that an "inertial" force $-ma_{\mathrm{com}}$ also acts on both masses.

 

[MattGeo]

I honestly don't think I can figure this out without further assistance and insight. It's been quite a long time since I took an advanced classical mechanics course. I just enjoy thinking about physics and made up this problem. Could you please prompt me with any further hints or important information?

 

[ergospherical]

In the centre of mass frame, you have two "independent" springs on either side of the centre of mass. (You do need to think about what the "effective" spring constants of these two springs are. How do springs "in series" combine?).

The right-hand mass is acted upon by spring tension, the external force $F$ and the inertial force $-ma_{\mathrm{com}}$. The left-hand mass is acted upon by spring tension and the inertial force $-ma_{\mathrm{com}}$.

In either case, the problem is just that of a mass on a spring, subject to some constant external force. That is probably a problem that you have solved before -- what is the motion?

If any of these sub-problems are unfamiliar, then it might be worth refreshing your memory on them before attempting a more complex combined problem.

 

[anuttarasammyak]

How much duration of time you keep pushing the system ? If we know it, we know both energy and momentum of the system which tell us center of mass motion and vibration around it. We may not be able to get the duration time in an easy way.

 

[anuttarasammyak]

Lagrangian of the system would be
$$L=\frac{1}{2}m_1\dot{x_1}^2+\frac{1}{2}m_2\dot{x_2}^2-\frac{1}{2}k(x_2-x_1-l)^2+Fx_1\theta(x_1)\theta(L-x_1)$$
where l=0.5m, F=10 newton and L=10m.

 

[PeroK]

I honestly don't think I can figure this out without further assistance and insight. It's been quite a long time since I took an advanced classical mechanics course. I just enjoy thinking about physics and made up this problem. Could you please prompt me with any further hints or important information?

What you should be able to do is generate a differential equation for the system, simply by applying Newton's laws.

If we let $X$ be the displacement of the larger mass, $M$; and, $x$ be the displacement of the smaller mass, $m$; and, $F$ be the externally applied force to the large mass; then, the spring force at any time is $k(X-x)$. This leads to the equations of motion:
$$ma_m \equiv m\ddot x = F_m = k(X-x)$$$$Ma_M \equiv M\ddot X = F - k(X-x)$$If we let $z = (X - x)$, then we have:
$$\ddot z = \ddot X - \ddot x = \frac F M - \frac k M(X-x) - \frac k m (X - x) = \frac F M - k(\frac 1 M + \frac 1 m)(X -x)$$Hence, we have an equation of motion for $z$:
$$\ddot z = \frac F M - k(\frac{M + m}{Mm})z$$And then take it from there.

 

[MattGeo]

What you should be able to do is generate a differential equation for the system, simply by applying Newton's laws.

If we let $X$ be the displacement of the larger mass, $M$; and, $x$ be the displacement of the smaller mass, $m$; and, $F$ be the externally applied force to the large mass; then, the spring force at any time is $k(X-x)$. This leads to the equations of motion:
$$ma_m \equiv m\ddot x = F_m = k(X-x)$$$$Ma_M \equiv M\ddot X = F - k(X-x)$$If we let $z = (X - x)$, then we have:
$$\ddot z = \ddot X - \ddot x = \frac F M - \frac k M(X-x) - \frac k m (X - x) = \frac F M - k(\frac 1 M + \frac 1 m)(X -x)$$Hence, we have an equation of motion for $z$:
$$\ddot z = \frac F M - k(\frac{M + m}{Mm})z$$And then take it from there.

great, thanks!! something as simple as you setting that up went a long way in dusting off all the cobwebs.

 

[MattGeo]

What you should be able to do is generate a differential equation for the system, simply by applying Newton's laws.

If we let $X$ be the displacement of the larger mass, $M$; and, $x$ be the displacement of the smaller mass, $m$; and, $F$ be the externally applied force to the large mass; then, the spring force at any time is $k(X-x)$. This leads to the equations of motion:
$$ma_m \equiv m\ddot x = F_m = k(X-x)$$$$Ma_M \equiv M\ddot X = F - k(X-x)$$If we let $z = (X - x)$, then we have:
$$\ddot z = \ddot X - \ddot x = \frac F M - \frac k M(X-x) - \frac k m (X - x) = \frac F M - k(\frac 1 M + \frac 1 m)(X -x)$$Hence, we have an equation of motion for $z$:
$$\ddot z = \frac F M - k(\frac{M + m}{Mm})z$$And then take it from there.

rustier than I thought. The differential equation immediately made sense to me. I just can't solve it because I don't remember anything from diff eq. It seems 2nd order linear and non-homogeneous but we can't integrate with respect to t. I assume I have to apply one of the standard techniques.

All that aside though, the more I think about the actual physics I am not sure how to figure out the problem even with the differential equation because we know X but not x. I was thinking I could just use the mass proportions and figure out how much the center of mass moved, given X, and then solve for x. I still can't parse the final kinetic energy of the system and the energy stored in the spring though after the 10 meter displacement of mass 2.

 

[PeroK]

rustier than I thought. The differential equation immediately made sense to me. I just can't solve it because I don't remember anything from diff eq. It seems 2nd order linear and non-homogeneous but we can't integrate with respect to t. I assume I have to apply one of the standard techniques.

All that aside though, the more I think about the actual physics I am not sure how to figure out the problem even with the differential equation because we know X but not x. I was thinking I could just use the mass proportions and figure out how much the center of mass moved, given X, and then solve for x. I still can't parse the final kinetic energy of the system and the energy stored in the spring though after the 10 meter displacement of mass 2.

Can you get another equation by considering the motion of the centre of mass? And the motion of the masses relative to the CoM?

 

[MattGeo]

the acceleration of the CoM is simply the applied external force divided by total mass of system. If I am in the CoM frame I am accelerating alongside it and it remains at rest with respect to me. I can't figure out how I would get the relative motion of the 2 masses with respect to CoM though. On the heavier mass there is the 10 N applied force on it and there is the spring force opposing that. But this is an accelerating reference frame, not an inertial frame, so shouldn't there be a fictious force to make everything balance out? I never thought this would be quite so difficult.

 

[PeroK]

the acceleration of the CoM is simply the applied external force divided by total mass of system. If I am in the CoM frame I am accelerating alongside it and it remains at rest with respect to me. I can't figure out how I would get the relative motion of the 2 masses with respect to CoM though. On the heavier mass there is the 10 N applied force on it and there is the spring force opposing that. But this is an accelerating reference frame, not an inertial frame, so shouldn't there be a fictious force to make everything balance out? I never thought this would be quite so difficult.

Regardless of the external force, the displacements of the masses relative to the CoM are related.

 

[TSny]

the acceleration of the CoM is simply the applied external force divided by total mass of system. If I am in the CoM frame I am accelerating alongside it and it remains at rest with respect to me. I can't figure out how I would get the relative motion of the 2 masses with respect to CoM though. On the heavier mass there is the 10 N applied force on it and there is the spring force opposing that. But this is an accelerating reference frame, not an inertial frame, so shouldn't there be a fictious force to make everything balance out? I never thought this would be quite so difficult.

Yes. In the CoM frame of reference, you must include a fictitious force (inertial force) for each mass as mentioned by @ergospherical. The fictitious forces are constant since the CoM has a constant acceleration relative to the inertial frame. So, they are easy to include in the analysis in the CoM frame.

You've chosen numerical values such that the spring is not stiff enough to prevent the 3 kg mass from colliding with the 1 kg mass. You'll find that $k$ needs to be greater than 10 N/m to avoid a collision (if you keep all other numerical values the same).

 

[MattGeo]

Regardless of the external force, the displacements of the masses relative to the CoM are related.

I've become extremely dense in time. I am at a complete dead end. I really want to learn how to solve this problem step by step because I think it is an excellent problem which encapsulates a lot of interesting physics concepts with some complicated behavior, despite the seemingly simple setup of the problem. I know that this site doesn't exist to simply solve problems for askers and people have to really learn by doing, but I just can't gather my thoughts and proceed without being babied and someone holding my hand through the procedure. It's just been too long.

 

[anuttarasammyak]

rustier than I thought. The differential equation immediately made sense to me. I just can't solve it because I don't remember anything from diff eq. It seems 2nd order linear and non-homogeneous but we can't integrate with respect to t. I assume I have to apply one of the standard techniques.

This differential equation for z seems that of forced harmonic oscillation. For initial condition of $z=0,\dot{z}=0$ for $t=0$,
$$z=\frac{F}{M\omega^2}(1-\cos \omega t)$$
where
$$\omega=\sqrt{k\frac{M+m}{Mm}}$$
$$\ddot{x}=\frac{k}{m}z=\frac{k}{m}\frac{F}{M\omega^2}(1-\cos \omega t)$$
It seems not dificult to integrate it with time twice to get displacement x and X=x+z-l as for position not displacement where l is natural length of spring. This is the situation for X < 10m.
We can get values of the parameters at X=10m to know P.E. and K.E. then and after.

 

[PeroK]

I've become extremely dense in time. I am at a complete dead end. I really want to learn how to solve this problem step by step because I think it is an excellent problem which encapsulates a lot of interesting physics concepts with some complicated behavior, despite the seemingly simple setup of the problem. I know that this site doesn't exist to simply solve problems for askers and people have to really learn by doing, but I just can't gather my thoughts and proceed without being babied and someone holding my hand through the procedure. It's just been too long.

I think @ergospherical is right. This problem is too advanced given your rustiness. You need to start with some simpler problems and build up your skills before tackling advanced problems. In this thread, all the serious efforts are from helpers.

 

[MattGeo]

This differential equation for z seems that of forced harmonic oscillation. For initial condition of $z=0,\dot{z}=0$ for $t=0$,
$$z=\frac{F}{M\omega^2}(1-\cos \omega t)$$
where
$$\omega=\sqrt{k\frac{M+m}{Mm}}$$
$$\ddot{x}=\frac{k}{m}z=\frac{k}{m}\frac{F}{M\omega^2}(1-\cos \omega t)$$
It seems not dificult to integrate it with time twice to get displacement x and X=x+z-l as for position not displacement where l is natural length of spring. This is the situation for X < 10m.
We can get values of the parameters at X=10m to know P.E. and K.E. then and after.

what exactly did you mean when you said "to get displacement x and X = x+z-l as for position not displacement" ?

 

[anuttarasammyak]

Lagrangian of the system would be
$$L=\frac{1}{2}m_1\dot{x_1}^2+\frac{1}{2}m_2\dot{x_2}^2-\frac{1}{2}k(x_2-x_1-l)^2+Fx_1\theta(x_1)\theta(L-x_1)$$
where l=0.5m, F=10 newton and L=10m.

$x_1,x_2$ are coordinates. Displacements are $X=x_ 1, x=x_2-l, z:=X-x=x_2-x_1-l$. We need to add/delete l for transformation between the two.
E.g. at t=0,
$$X=0,x=0,x_1=0,x_2=l$$