Astronaut at centre of tetrahedron

[Saitama]

Homework Statement

An astronaut in the International Space-station attaches himself to the four vertices of a regular tetrahedron shaped frame with 4 springs. The mass of the springs and their rest length are negligible, their spring constants are ##D_1=150 N/m##, ##D_2=250 N/m##, ##D_3=300 N/m## and ##D_4=400 N/m##. For the sake of simplicity - the astronaut is considered pointlike, its mass is m=70 kg. What is the period of the oscillation of the astronaut if he is displaced from his equilibrium position and released?

Homework Equations

The Attempt at a Solution

I haven't yet attempted the problem by writing down the equations when the astronaut is displaced from centre because looking at the answer given, I think there is a shorter way. The given answer is

$$2\pi \sqrt{\frac{m}{D_1+D_2+D_3+D_4}}$$​

Looking at the answer, it is similar to the time period of mass hanging by four parallel springs. Is this by coincidence or is there a reason behind this? I am clueless if there is a reason.

If it is a coincidence, I am ready to make the equations.

Any help is appreciated. Thanks!

 

[consciousness]

Direction of displacement isn't given?

 

[Saitama]

Direction of displacement isn't given?

I checked the problem again. I have copied the problem word by word. :)

 

[voko]

I can't think of any clever way.

 

[I like Serena]

This can't be right.


Consider 1 spring.
Energy conservation tells us that ##\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 = C##.
This yields the period ##2\pi \sqrt{\frac m {D_1}}##.


Now consider 2 opposite springs.
A deviation in the direction of the springs gives:
$$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 x^2 = C$$
$$\frac 1 2 m \dot x^2 + \frac 1 2 (D_1 + D_2) x^2 = C$$
This is equivalent to 1 spring with constant ##D_1 + D_2##.
At least this goes into the direction of the answer.


Now consider 3 springs in a equilateral triangle with equal spring constants ##D_1=D_2=D_3## and a small deviation into the the direction of the first spring:
$$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 (\frac 1 2 x)^2 + \frac 1 2 D_3 (\frac 1 2 x)^2 = C$$
$$\frac 1 2 m \dot x^2 + \frac 1 2 (D_1 + \frac 1 2 D_2 + \frac 1 2 D_3) x^2 = C$$
The resulting period is equivalent to a spring with constant ##D_1 + \frac 1 2 D_2 + \frac 1 2 D_3## and not a spring constant of ##D_1 + D_2 + D_3##.


The same principle applies to 4 springs in a tetrahedron.

 

[TSny]

Interesting. I find that the restoring force does not depend on the direction of the displacement from equilibrium and the restoring force obeys Hooke's law even for large displacements. The effective force constant has a simple relationship to the individual spring constants.

 

[TSny]

Looking at the answer, it is similar to the time period of mass hanging by four parallel springs. Is this by coincidence or is there a reason behind this? I am clueless if there is a reason.

Consider the general expression for the potential energy of the system for an arbitrary location of the astronaut. Keep in mind that you are to neglect the natural length of each spring.

 

[haruspex]

A deviation in the direction of the springs gives:
$$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 x^2 = C$$

That would be true if the springs were initially at their relaxed lengths. Here, they're to be taken as effectively zero relaxed length.

 

[I like Serena]

That would be true if the springs were initially at their relaxed lengths. Here, they're to be taken as effectively zero relaxed length.

Good point.
The result for 2 springs remains the same though.

 

[haruspex]

Consider the general expression for the potential energy of the system for an arbitrary location of the astronaut.

It's quite easy just using position and force vectors.
If v_i is the vector position of one vertex, x the vector of the astronaut at equilibrium position, and Δx a displacement vector, what is the vector for the force towards that vertex?

 

[TSny]

It's quite easy just using position and force vectors.
If v_i is the vector position of one vertex, x the vector of the astronaut at equilibrium position, and Δx a displacement vector, what is the vector for the force towards that vertex?

Yes! It does fall out almost immediately with forces and position and displacement vectors.

I wanted to avoid having to think about vectors.

 

[Saitama]

Now consider 2 opposite springs.
A deviation in the direction of the springs gives:
$$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 x^2 = C$$
$$\frac 1 2 m \dot x^2 + \frac 1 2 (D_1 + D_2) x^2 = C$$
This is equivalent to 1 spring with constant ##D_1 + D_2##.
At least this goes into the direction of the answer.

That would be true if the springs were initially at their relaxed lengths. Here, they're to be taken as effectively zero relaxed length.

Hi ILS and haruspex!

What is wrong with ILS method? haruspex, I am unable to interpret "they're to be taken as effectively zero relaxed length.".

Consider the general expression for the potential energy of the system for an arbitrary location of the astronaut. Keep in mind that you are to neglect the natural length of each spring.

It's quite easy just using position and force vectors.
If v_i is the vector position of one vertex, x the vector of the astronaut at equilibrium position, and Δx a displacement vector, what is the vector for the force towards that vertex?

The force vector towards the vertex is along Δx-x+v_i, is this correct?

If the above is correct, the force vector is:
$$k\left(\left|\Delta \vec{x}-\vec{x}+\vec{v_i}\right|-\left|\vec{x}-\vec{v_i}\right|\right)\frac{\Delta \vec{x}-\vec{x}+\vec{v_i}}{\left|\Delta \vec{x}-\vec{x}+\vec{v_i}\right|}$$
where k is the spring constant.

 

[voko]

The force vector towards the vertex is along Δx-x+v_i, is this correct?

Let the non-equilibrium position be ##\vec X##. Then the force due to the i-th spring must be directed from ##\vec X## to ##\vec v_i##. Is that ##\vec X - \vec v_i## or ##\vec v_i - \vec X##?

Now back to the equilibrium position ##\vec x## and the displacement ##\Delta \vec x##. What is ##\vec X## in these?

If the above is correct, the force vector is:
$$k\left(\left|\Delta \vec{x}-x+v_i\right|-\left|x-v_i\right|\right)\frac{\Delta \vec{x}-x+v_i}{\left|\Delta \vec{x}-x+v_i\right|}$$
where k is the spring constant.

Note that the relaxed length is negligible. What is the elongation of the i-th spring in this case?

 

[Saitama]

Let the non-equilibrium position be ##\vec X##. Then the force due to the i-th spring must be directed from ##\vec X## to ##\vec v_i##. Is that ##\vec X - \vec v_i## or ##\vec v_i - \vec X##?

It's the latter, sorry, I made a dumb mistake.

Now back to the equilibrium position ##\vec x## and the displacement ##\Delta \vec x##. What is ##\vec X## in these?

##\vec X## is ##\vec x + \vec {\Delta x}##.

Note that the relaxed length is negligible. What is the elongation of the i-th spring in this case?

I am not getting this, why is the relaxed length negligible?

 

[voko]

I am not getting this, why is the relaxed length negligible?

Because the problem said so.

 

[Saitama]

Because the problem said so.

The initial length of spring given by ##|\vec{x}-\vec{v_i}|## is negligible hence the extension in spring is given by ##|\vec{x}+\vec{\Delta x}-\vec{v_i}|##, correct?

 

[voko]

The initial length of spring given by ##|\vec{x}-\vec{v_i}|##

I am not sure what "initial" means here, but Hooke's law deals with the extension from the relaxed length. ##|\vec{x}-\vec{v_i}|## is the length at equilibrium, not the relaxed length.

the extension in spring is given by ##|\vec{x}+\vec{\Delta x}-\vec{v_i}|##, correct?

Yes.

 

[Saitama]

I am not sure what "initial" means here, but Hooke's law deals with the extension from the relaxed length. ##|\vec{x}-\vec{v_i}|## is the length at equilibrium, not the relaxed length.

Yes.

Do I have to write the total potential energy of the springs?

The potential energy in the springs when the astronaut is displaced from equilibrium position is given by:
$$\sum_{i=1}^4 \frac{1}{2}D_i |\vec{x}+\vec{\Delta x}-\vec{v_i}|^2$$
Do I just expand the expression?

 

[voko]

As haruspex suggested, it is probably best to stay with forces. Find the resultant.

 

[Saitama]

As haruspex suggested, it is probably best to stay with forces. Find the resultant.

The resultant force is given by:
$$\vec{F}=\sum_{i=1}^4 D_i\vec{v_i}-\vec{x}\sum_{i=1}^4 D_i -\vec{\Delta x}\sum_{i=1}^4 D_i$$

The above equation is of SHM with angular frequency ##\sqrt{D_1+D_2+D_3+D_4}##. This is the correct answer. Thanks a lot everyone for the help.

So, there is no relation between this problem and the case of mass hanging by 4 parallel springs, correct?

 

[haruspex]

The initial length of spring given by ##|\vec{x}-\vec{v_i}|## is negligible hence the extension in spring is given by ##|\vec{x}+\vec{\Delta x}-\vec{v_i}|##, correct?

and from post #12

The force vector towards the vertex is along Δx-x+v_i

So putting those together, what is the force vector?

 

[voko]

So, there is no relation between this problem and the case of mass hanging by 4 parallel springs, correct?

It is instructive to look back at the problem and find out why the result is that way.

Is it because of the tetrahedron?

Is it because of the ISS environment?

 

[Saitama]

Is it because of the tetrahedron?

Is it because of the ISS environment?

I really have no clue. :(

 

[voko]

Did you use the tetrahedron's symmetries in your derivation? Did it in fact matter that there were four springs and not twenty four?

How did the ISS thing affect your derivation? What does that even mean? What if whatever that means was removed, would the result change?

 

[Saitama]

Did you use the tetrahedron's symmetries in your derivation?

Nope.

Did it in fact matter that there were four springs and not twenty four?

No.

How did the ISS thing affect your derivation?

It did not affect the derivation.

What does that even mean?

I am not sure, I guess that means we can make any arbitrary kind of connection of springs and the result (for time period) we get is always the same.

What if whatever that means was removed, would the result change?

The result won't change.

Am I correct?

 

[voko]

You did not use the tetrahedron's symmetries and the number of springs, which suggests the result should be independent of that. Yet there is a subtlety. Assume there is just one spring. It can obviously oscillate. But there is another periodic motion which is possible - what is it? What about two and more springs? What is the condition to eliminate such motion?

As for the ISS condition, I do not understand what you said there.

 

[Saitama]

You did not use the tetrahedron's symmetries and the number of springs, which suggests the result should be independent of that. Yet there is a subtlety. Assume there is just one spring. It can obviously oscillate. But there is another periodic motion which is possible - what is it? What about two and more springs? What is the condition to eliminate such motion?

I can't see the second possible motion for the case of 1 spring but I can think of two possible motions in case of two springs.

As for the ISS condition, I do not understand what you said there.

I meant that even if the tetrahedron was not kept in ISS, we would still get the same result. Is this what you asked me or did I misinterpret your question?

 

[voko]

In principle, any string is also a spring (of great stiffness). What kind of motion is possible for a mass on a string?

Regarding ISS, I did not see your explanation on what effect that introduced or eliminated. What if the system were on the Earth? Would you need to take (or not take) something else into account?

 

[Saitama]

In principle, any string is also a spring (of great stiffness). What kind of motion is possible for a mass on a string?

Simple harmonic and circular?

Regarding ISS, I did not see your explanation on what effect that introduced or eliminated. What if the system were on the Earth? Would you need to take (or not take) something else into account?

I don't think I will have to take anything else into account. Also, I think the rest lengths of the springs would change on Earth but that won't affect the final result, correct?

 

[voko]

Simple harmonic and circular?

So what would need to be done to prevent circular motion?

I don't think I will have to take anything else into account. Also, I think the rest lengths of the springs would change on Earth but that won't affect the final result, correct?

Would you not need to consider the pull of gravity?

 

[Saitama]

So what would need to be done to prevent circular motion?

I would displace the mass slightly along the string to make it perform SHM, but I am not sure why you ask me to "prevent" circular, I simply won't give it enough velocity perpendicular to string for performing circular motion.

Would you not need to consider the pull of gravity?

That comes in the force expression but that won't affect the angular frequency. :)

 

[haruspex]

In principle, any string is also a spring (of great stiffness). What kind of motion is possible for a mass on a string?

Need to be a bit careful there. One attribute that was definitely critical to the neat result is the zero relaxed length of the springs. A single such of great stiffness would therefore be of very short length. [STRIKE] I can see where you are going, and your argument is sound, but keep the stiffness moderate.[/STRIKE]
Edit:
The zero relaxed length does invalidate the argument. Back to the equations:
net force = ##\Sigma D_i(\vec{v_i} - \vec{x} - \vec{\Delta x}) + m \vec{g}##
At equilibrium:
0 = ##\Sigma D_i(\vec{v_i} - \vec{x}) + m \vec{g}##
whence
net force = ##-\vec{\Delta x}\Sigma D_i##

 

[voko]

I would displace the mass slightly along the string to make it perform SHM, but I am not sure why you ask me to "prevent" circular

In certain spring arrangements you have rotary motion in addition to oscillations. The original problem used the tetrahedron arrangement to prevent rotary motion and my question is what is the general condition on the arrangement to prevent rotary motion.

 

[voko]

Need to be a bit careful there. One attribute that was definitely critical to the neat result is the zero relaxed length of the springs. A single such of great stiffness would therefore be of very short length. [STRIKE] I can see where you are going, and your argument is sound, but keep the stiffness moderate.[/STRIKE]
Edit:
The zero relaxed length does invalidate the argument.

I have lost you here. What argument is invalidated?

 

[haruspex]

I have lost you here. What argument is invalidated?

Maye I misguessed where you were going, but it seemed like you were heading towards concluding that the result would not apply in a gravitational field. The equations show that it does.
If I interpreted correctly, the flaw in your argument (treating a string as a very stiff spring) is that a spring of arbitrarily great stiffness and zero relaxed length would always have zero length.