# Spring with 2 masses free fall

1. Jan 3, 2014

### R136a1

1. The problem statement, all variables and given/known data

3. The attempt at a solution

The reference to exercise 1.7 is not essential. The only part exercise 1.7 that is relevant to this exercise is the following: if a long narrow tube is drilled between antipodal points on a sphere of uniform mass density and two identical masses are dropped one after the other then there is a tidal force $F = fx$ between the two masses where $f$ is as given in the above exercise and $x$ is the separation between the two masses.

Also I'm 99.99% sure that the problem statement has a typo. If this were a system of two identical masses attached together by a massless spring of spring constant $k$ with no external forces present then the period would be $T = 2\pi \sqrt{\frac{m}{2k}}$ and not what is given in the problem statement.

Anyways, I approached this problem similarly to how one approaches the problem of two identical masses attached by a massless spring with no external forces present: by going to the center of mass frame and working out the equations of motion of the identical masses in this frame. Let $l$ be the unstretched length of the spring, $r_a$ the position vector from the center of mass to the top mass, and $r_b$ the position vector from the center of mass to the bottom mass; we take the up direction (i.e. the direction of $r_a$) to be the positive one. Note that $r_a - r_b$ gives the instantaneous length of the spring hence $r_a - r_b - l$ is the displacement of the spring from its unstretched length.

Now the center of mass is itself a hypothetical particle that's also in free fall in this scenario. Hence in this frame it sees a tidal force $f r_a$ pushing the top mass away from it and a tidal force $f r_b$ pushing the bottom mass away from it since this is exactly what the longitudinal tidal force is physically. Hence the equation of motion of the top mass will be $m \ddot{r}_a = -k(r_a - r_b - l) + fr_a + ma$ and that of the bottom mass will be $m \ddot{r}_b = k(r_a - r_b - l) - fr_b + ma$ where $+ma$ is the upwards inertial force due to the downward free fall of the center of mass frame.

Subtracting the two equations we have $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l)+ f(r_a + r_b)$ which is a problem because $r_a +r_b = 0$ since this is the center of mass frame. But this reduces to the equation one would normally get in the center of mass frame for a free system consisting of a spring with two masses, the resulting natural period of which is $T = 2\pi \sqrt{\frac{m}{2k}}$, so I definitely went wrong somewhere but can't figure out where.

2. Jan 3, 2014

### Staff: Mentor

I agree.

The tidal forces are relative to the center of mass. You have to multiply f with that distance, so both masses should get the same sign in front of fri.

3. Jan 3, 2014

### R136a1

Thanks for the reply! As for the tidal forces having the same sign: if we imagine the center of mass as a hypothetical particle relative to whom both the top and bottom mass experience tidal forces then shouldn't both the top and bottom mass be accelerating away from the center of mass under action of the tidal forces, by definition of the longitudinal gravitational tidal force? Hence relative to the center of mass shouldn't the bottom mass receive a force downwards from the tidal force and the top mass a force upwards from the tidal force, thus having a difference in sign? Thanks again.

4. Jan 4, 2014

### haruspex

No, I agree with the expression given for the period. It's the same as for a single mass attached to a fixed point by a spring half as long but with the same constant. Isn't that 2π√(k/m)?
Which mass experiences the greater gravitational force?

5. Jan 4, 2014

### Staff: Mentor

The difference in sign is already included in the different signs for the ri.

@haruspex: 2π√(m/k) should be the formula for a single mass, attached to a fixed point with a spring of constant k.
$m\ddot{r} = -k r$ => $r(t)=A sin(\omega t +\theta)$ where $\omega^2 = \frac{k}{m}$
Therefore, the period is $\frac{2 \pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$

6. Jan 4, 2014

### R136a1

I'm not sure I follow, at least with regards to the physical system in this problem. We have a spring with two identical masses one on each end and if the system is allowed to oscillate freely then the period should be $T = 2\pi \sqrt{\frac{m}{2k}}$.

See for example ehild's post here: https://www.physicsforums.com/showpost.php?p=2507137&postcount=11

Whilst falling through the tunnel, the bottom mass would feel a greater gravitational force up until the zero point of the oscillatory motion through the tunnel (due to the gravitational force-not the oscillatory motion of the spring) wherein the gravitational force switches direction and pulls the mass-spring system back upwards and now the top mass would feel a greater gravitational force. At least that's how I think it should work.

Ah right I totally forgot that $r_b = -r_a$ so the sign difference is implicit in the center of mass position vectors. Sorry about that.

I hope my thought process is at least somewhat correct: the spring with two masses on either end is dropped into the tunnel. The longitudinal gravitational tidal force will immediately pull the two masses apart and this will have the spring force pulling them in until the turning point wherein it pushes them out Ad infinitum. So the tidal force acts as like a driving force for the oscillatory motion of the spring.

The tidal force can be included as a force in the equations of motion if we go to the reference frame of one particle relative to whom the other particle seems to be accelerating away due to the tidal force (after taking into account inertial forces) because physically this is how the tidal force will manifest itself in the equations of motion, being that it represents the relative acceleration between two particles. Unfortunately, as far as I'm aware, a spring with two masses is best solved in the center of mass frame and not the frame of either of the two endpoint masses. And since we can think of the center of mass as a hypothetical particle (which is also in free fall in this scenario), we can think of the tidal forces between the center of mass particle and each of the two endpoint masses attached to the spring so as to include them in the equations of motion (after taking into account the inertial force in the freely falling center of mass frame).

After taking into account the implicit sign difference, I get $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l) + f(r_a - r_b)$ which still doesn't seem to be what I need. Shouldn't I be getting something like $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l) - f(r_a - r_b - l)$ in order to conclude that the new period is $T = 2\pi \sqrt{\frac{m}{2k + f}}$?

Thanks again for the help haruspex and mfb.

7. Jan 4, 2014

### Staff: Mentor

No, your equation (the first one) is fine, and the result is right as well. The equilibrium point shifts a bit (so the spring is extended a bit if there is no oscillation - that does not influence the period), and the effect on the period is like a modified spring constant.

8. Jan 4, 2014

### R136a1

Thank you so much mfb. If we let $u = r_a - r_b - l$ in the equation $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l) + f(r_a - r_b)$ then we get $m\ddot{u}+ (2k - f)u = fl$. Wouldn't this imply that the new period is $T = 2\pi \sqrt{\frac{m}{2k - f}}$ as opposed to $T = 2\pi \sqrt{\frac{m}{2k + f}}$? As far as I can tell, the only way it can be $T = 2\pi \sqrt{\frac{m}{2k + f}}$ is if we had something like $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l) - f(r_a - r_b)$ but this would imply that the tidal force itself drives a sinusoidal motion between the freely falling masses i.e. $m(\ddot{r}_a - \ddot{r}_b) = - f(r_a - r_b)$, ignoring the spring force, which doesn't make sense since the tidal force acts to make the separation between two freely falling masses larger and larger as time goes on. Also doesn't $T = 2\pi \sqrt{\frac{m}{2k - f}}$ make more sense than $T = 2\pi \sqrt{\frac{m}{2k + f}}$? The tidal force continuously increases the separation between the masses attached to the spring as time goes on which would make it so that the mass-spring system takes a longer time to go from amplitude to amplitude in its overall sinusoidal motion, thus having a larger period as opposed to a smaller period.

9. Jan 4, 2014

### haruspex

You're right - in thinking of it as a single mass with a spring of half the length attached to a fixed point, I should have doubled the spring constant.

10. Jan 4, 2014

### haruspex

Why? How does the force vary as distance from the centre of the earth when inside the earth?

11. Jan 4, 2014

### R136a1

Oh sorry I meant to say the opposite of what I actually said. I was thinking of the field external to the sphere of uniform density. Inside it varies linearly with the distance from the center of the sphere of uniform density.

12. Jan 4, 2014

### haruspex

Right, so does the tidal force contribute positively or negatively to the acceleration of the distance between them? The answer to that should tell you whether it is 2k+f or 2k-f.

13. Jan 4, 2014

### R136a1

Thank you very much haruspex! I was doing the entire problem with the mindset of someone analyzing an inverse-square gravitational field such as that exterior to a sphere of uniform density, although I have absolutely no idea why I was stuck in this mindset you'll have to excuse the stupid mistake!

Ok so starting from the top (no pun intended :tongue2:), we go to the center of mass frame of the mass-spring system that's in free fall inside the sphere of uniform density. As before the direction pointing from the center of mass to the top mass is positive and the direction pointing from the center of mass to the bottom mass is negative. The position vectors are as before $r_a$ for the top mass and $r_b$ for the bottom mass, with the respective signs implicitly absorbed into the respective position vectors.

Because the gravitational force inside scales linearly with the distance of a given mass from the center of the sphere, the top mass accelerates faster than the center of mass and the center of mass accelerates faster than the bottom mass. Hence the center of mass attributes an attractive gravitational tidal force to both of the masses at the ends of the spring.

The resulting equations of motion are then $m\ddot{r}_a = -k(r_a - r_b - l) - fr_a + m a$ and $m\ddot{r}_b = k(r_a - r_b - l) - fr_b + ma$ where ma is the upwards inertial acceleration due to the downwards free fall acceleration of the center of mass, so the combined equation of motion is $m(\ddot{r}_a - \ddot{r}_b) = -2k(r_a - r_b - l) - f(r_a - r_b)$; letting $u = r_a - r_b - l$ we then get $m\ddot{u} + (2k + f)u = -fl$. The $-fl$ term just shifts the equilibrium point of the mass-spring system and we get the desired period $T = 2\pi\sqrt{\frac{m}{2k + f}}$.

Does that look fine? And thanks again to both of you.

Last edited: Jan 4, 2014
14. Jan 4, 2014

### Staff: Mentor

Looks good.

Interesting fact: this period is the same as the overall period of the free-falling mass through earth. This is easy to see for two objects in the center, but their relative tidal motion is independent on the position as long as they are inside earth.

This all assumes a constant density of earth, of course, which is not true. But the tunnel is purely hypothetical as well, so who cares...