# Pendulum-Spring System in SHM

1. Dec 23, 2005

### Doc37

Hey,
I have a debate that needs to be resolved. Here is the diagram of the problem and I will describe below:

It is essentially a string that has a spring attached to the bottom with a weight of mass m attached to the spring. This is system is pulled back and released. Is it possible that if the period of both spring and the pendulum in simple harmonic motion are matching, that the weight of mass m will travel linearly, i.e. the spring will compress and expand at a rate that the weight is neither raised nor lowered by the system. Is this possible even on a theoretical level? When I work think about it, it seems like it should be possible to get a spring and a pendulum to both oscillate in such a way that the weight travels in a straight line. Maybe not, but I thought I would run it by some experts to get your opinions. If you need more information just respond to this letting me know and I can answer any questions you might have.
Thanks in advance for any help.

2. Dec 23, 2005

### Integral

Staff Emeritus
My intuition says no. Since the mass remains at a constant potential energy, there is no exchange between potential and kinetic energy to drive the system. This mode of motion could not be sustained.

3. Dec 24, 2005

### daniel_i_l

And, at "time 2" the spring should be more streched than at times 1 and 3 because at time 2, more centripital force is in the direction of the spring.

4. Dec 24, 2005

### krab

I find that under the condition that the spring constant exactly equals the weight of the mass divided by the length from mass to pivot (when mass is at the centre of its stroke), then vertical forces exactly cancel and there is no up-down movement.
In answer to Integral, the potential energy is in the spring, no longer in the gravitation.
Also, the diagram is wrong, since the spring will remain in line with the string.

5. Dec 24, 2005

### lightgrav

I was concerned that the motion might not be close to simple harmonic -
but the acceleration goes as tan(theta) , so it's pretty good.

For the spring Force's vertical component to equal Earth's gravity Force,
then F_spring cos(theta) = ks cos(theta) = mg .
Modelling the spring as a Hooke's law spring with relaxed length R,
the mass is at (R+s) cos(theta) = L (constant) below the pivot,
so the stretch must be s = L/cos(theta) - R = (mg/k)/cos(theta).
For this to be valid at all theta, R must be zero, and k = mg/L .

A relaxed length of zero can be simulated by putting the mass on a string,
guided over a pulley to the spring, which is relaxed when the mass is at the pivot.

m a = F_spring sin(theta) = -(mg/cos(theta)) sin(theta) = - mg tan(theta) .

Daniel_i_l : a straight-line path does not curve around any "center",
so the direction of a center is undefined; can't take Force components in the direction of it. The closest you can do is to look at Force components
perpendicular to the velocity (that is, vertical Force components).

6. Dec 25, 2005

### daniel_i_l

Wait a minute, from your replies I see that you think that the spring is on the same line as the string. In the picture it looks like the spring was held from the top and released with the spring facing down.
So, in the "time 1" position, only gravity pulls the spring down and
Fh - mg =0,
but as it gets to "time 2" the spring is hanging is line with te string and:
Fh - mg = 1/2mv^2
any way during the whole trip there's atleast mg pulling on the string so how could it ever get shorter than it was at "time 1" ?

7. Dec 25, 2005

### Gamma

Even if it was released in this fasion, in 'time3' the spring and the string should align unlike shown in the picture.

Still we can ask the question, between stage1 and stage 2, will the mass m remain horizontal. If you consider the mass and the spring as one whole unit, say mass M, it is now like a simple pendulum with a mass M at its end. What upward force is acting on 'M' which result in the comression of the spring? we know mg is downward. I don't see any upward forces. Therefore I believe the mass level will be lower than that of stage 1.

It is not hard to do this experiment. You need to choose a spring of known k and choose the string length such that the periods match as stated in the problem. Would be interesting.

Gamma.

8. Dec 25, 2005

### Jelfish

Well, the equation of motion for the pendulum is:

$$\ddot{\theta}+\omega _0 ^2 \sin{\theta} = 0$$

so the solutions aren't simple sinusoidal unless you take the small angle approximation. So maybe it would work for small angles, but even then it wouldn't be exact.

Last edited: Dec 25, 2005
9. Dec 25, 2005

### daniel_i_l

I thought a bout it some more and now I'm sure that it can't happen. For one thing - the energy problem - for the spring PE to turn into KE, the spring has to be less compressed when it gets to "time 2", this is impossible, there is no force that would "push" the mass up, if anything the greater tension at "time 2" would result with the spring being streched more, not less. Also, as mentioned, at "time 3" the spring would be in the same direction as the string, this means that for the mass to be at the same height the spring has to be longer tham it was at "time 1" - this clearly contridicts energy conservation.

10. Dec 26, 2005

### krab

That's for an ordinary pendulum whose length is constant. That's not the case here.
Say the length at centre is L, and when it is at angle $\theta$, it is longer by length e. Say the spring constant is k, so this adds a force ek in the direction along the pendulum. We want to arrange things so the resultant vertical force on the bob is zero, so it only moves horizontally. The total force is $F_0+ek$, where $F_0$ is the force when the pendulum is un extended, namely mg. Then the vertical force equation is
$$mg=(mg+ek)\cos\theta$$
But from the geometry, it is clear that
$$\cos\theta={L\over L+e}$$
Plug this into equation above, and you find e cancels:
$$mg=kL$$
The fact that e cancels means that this balancing out of the vertical force will continue even as e changes with time. Which is what you want for the motion to continue horizontal.QED

11. Dec 26, 2005

### Jelfish

woops. thanks for the explanation

12. Dec 26, 2005

### daniel_i_l

Oh krab, from you explanation I see that at time 1 the spring is streched past its equilibrium, if that is true than I see how it was possible, I was assuming that it was released from rest. I also see that unlike the picture, it is possible if the spring is released when it is on the same line as the string

13. Dec 26, 2005

### andrevdh

In the drawing you can see that the tension tends to rotate the mass, both at points one and two in the motion, around the com, in such a manner that the point of application of the tension and the com are brought in line with each other.

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14. Dec 27, 2005

### andrevdh

In attachment 1 we are looking at the motion of the mass along the line of the string. It is experiencing two forces along the direction of the string, the $W\cos(\theta)$ component of its weight and the restoring force $R$ of the spring. This weight component varies between a minimum $W_{min}$ and maximum $W_{max}$ as the pendulum swings between its amplitudes $\theta_A \ and \ -\theta_A$ as shown in attachment 2. As the component of the weight therefore varies between its minimum and maximum values the mass will therefore oscillate along the line of the string with amplitude $A$. The pendulum will therefore be at its longest when at $\theta=0$ with a lenght of $l+A$ and at its shortest when at $\theta_A \ and \ -\theta_A$ with lenght $l-A$ - see attachment 1. The position of the mass as it swings to and fro will therefore be as showm in attachment 3. Clearly it is not possible for it to stay on the same level during its swing.

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Last edited: Nov 29, 2006
15. Dec 28, 2005

### andrevdh

The mentioning of amplitudes A in my previous post implies that the motion of the mass along the line of the string is SHM, something that I did not investigate. Therefore one should rather think of a single extension of the spring from the the one extreme position to the other. The rest I stand by unless someone is so kind as to convince me otherwise.

16. Jan 9, 2006

### Doc37

Wow, I disappear due to the holidays and a big discussion ensues. I appreciate everyones input into this question. From what I see, there are two differing opinions between krab and andrevdh. Looking at both of the explanations krab makes the most sense to me. I find it interesting though because it is two different approaches to the same problem and it seems that two different answers resulted. Any thoughts from anyone on who they think makes the most sense?

17. Jan 9, 2006

### andrevdh

The problem I have with krab's remark that the total force is $F_o+ek$ is that he is forcing the pendulum to get longer as it swings up from $\theta=0$ (which is what is required if the mass is to stay on the same height), but will this be the case in real life? As I see it the spring responds to the component $W\cos(\theta)$ of the weight. As this component decreases as the pendulum swings up from the bottom the spring will respond by contracting, not extending.

18. Jan 9, 2006

### lightgrav

The Tension in a spring depends on its stretch (it does not "respond to" external Force components except via its stretch).

The original post asked whether horizontal motion was possible, which is what krab and I have answered. There are conditions on the apparatus (must arrange for spring with zero effective length) and on initial conditions (v=0 at L = mg/k), which have been mentioned in the posts.
It is most straight-forward to analyze straight-line using rectangular coordinates, not circular coordinates.

There are certainly other motions of a mass-on-spring that are possible!
Hang a mass from a long spring, bouncing around its vertical equilibrium.
Adjust the length (with string, if necessary) until a small-angle swing angular frequency $$\omega_h$$ is ½ of the vertical angular frequency $$\omega_v$$ .
Now displace the mass by a small angle, and a small distance downward, before letting go. The mass will peak at the center, and have lowest heights at each end of the swing.
Or displace by a small angle, and a small distance upward, before letting go. The mass will peak at the ends of the swing, and be lowest in the center.
The horizontal path asked about in this thread is, in a sense, halfway between these two initial conditions.

Last edited: Jan 9, 2006
19. Jan 11, 2006

### andrevdh

Oh, I see the motion have much more interesting possibilities than I originally thought. Another mistake I made is reasoning with Newton's laws in a noninertial frame of reference - the motion along the line of the pendulum is in an accelerating system.

20. Jan 19, 2006

### Doc37

Wow these conditions are however tough to replicate in real life. I tried to do it last weekend but i couldnt get it to work. Really close a couple of times but nothing exact, it seems to me that period definitely plays into it, the couple of times it seemed good it would all of a sudden become out of whack in the 2nd, 3rd, 4th, etc. passes. You guys have any insight on this problem?