# What is the significance of clamping the center of the spring?

7. A block is hung on a spring, and the frequency $f$ of the oscillation of the system is measured. The block, a second identical block, and the spring are carried in the Space Shuttle to space. The two blocks are attached to the ends of the spring, and the system is taken out into space on a space walk. The spring is extended, and the system is released to oscillate while floating in space. What is the frequency of oscillation for this system, in terms of $f$?

Q15.7 We assume that the coils of the spring do not hit one another. The frequency will be higher than $f$ by the factor $\sqrt{2}$. When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move. We can imagine clamping the center coil in place without affecting the motion. We can effectively duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth. The half-spring with its center coil clamped - or its other half cut off - has twice the spring constant as the original uncut spring, because an applied force of the same size would produce only one-half the extension distance. Thus the oscillation frequency in space is $\biggl(\dfrac{1}{2\pi}\biggr)\biggl(\dfrac{2k}{m}\biggr)^{1/2} = \sqrt{2}f$. The absence of a force required to support the vibrating system in orbital free fall has no effect on the frequency of its vibration.

Can someone explain to me:

1. the role of clamping the center of the coil?
2. what is its relation with space, gravity and SHM?
• Although I usually agree with @dmckee on close worthy questions, I disagree on this one. I was composing an answer with the aim of drawing a parallel between the solution given and the (inverse of the) method of images. In other words, I think this question has a strong conceptual component, i.e., "why and/or how does this approach work?". Questions about solution methods and approaches are, I think, good questions in general. – Alfred Centauri Oct 15 '14 at 22:56
• i agree @dmckee to limit the questions as if they are HW but how to ask about a problem in a text book that is vague at least for me ? – Mohamed Osama Oct 15 '14 at 23:02
• @AlfredCentauri That's an elegant approach to explaining the solution, and perhaps I am in the wrong here, but at this point we haven't the slightest indication of what the OP has does or does not understand and a complete lack of identification of the "specific physics concept" at question. – dmckee --- ex-moderator kitten Oct 15 '14 at 23:02
• @MohamedOsama This could be saved if you could explain what you don't understand. There are several parts to the explanation given. Surely there are some bits you get and some you don't. If you don't get any of it you would be well advised to look at a simple question first. – dmckee --- ex-moderator kitten Oct 15 '14 at 23:10
• I've edited the question. Please keep in mind that our homework tag applies to many types of questions that are not actually homework, and in particular it does need to be on this question. (The name is misleading; we're working on that.) – David Z Oct 16 '14 at 1:50

i do not understand the part of clamping the center of the coil or cutting it off

The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer.

If you have taken a first semester in electrostatics, you may be aware of the method of images. In that case, there isn't a symmetry but we create a symmetric problem with a known solution by extending the domain with a mirror image.

This is essentially the reverse of that approach. We have a symmetric problem which we solve by appeal to the known solution of the unsymmetric problem by cutting the domain in half.

Since the problem is symmetric, we know that if one mass is displaced by $\Delta x$, the other mass is displaced by an equal and opposite amount.

Thus, the length of the spring changes by $2\Delta x$ so the magnitude of the force on either mass changes by $2k\Delta d$

The fact is, and this shouldn't be hard to convince yourself of, that if we only look at 'one side' of this problem, it appears that either mass is attached to a spring that is fixed at the center and has spring constant $2k$. Thus, the frequency of oscillation of this 'one sided' problem is

$$\omega = \sqrt{2k/m} = \sqrt{2}\sqrt{k/m}$$

Appealing to symmetry is a powerful method to 'intuit' an answer from known solutions.

As regards to the effect of gravity, assuming that the spring is linear, the frequency of oscillation is unaffected by a constant offset force.

That is to say, the frequency of oscillation, $\omega = \sqrt{2k/m}$ is the same for the following two differential equations:

$$m\ddot x + 2kx = 0$$

$$m\ddot x + 2kx = mg$$

The effect of the constant gravitational force on the right hand side of the 2nd equation is essentially to add a constant offset to the solution for the position function without changing the frequency.

The equation of motion for one mass $m$ attached to a spring with spring constant $k$ (with the other end fixed) on Earth is $$\ddot{x} + \omega^2 x = g$$ where $2 \pi f = \omega = \sqrt{k / m}$. If it's not obvious that the $g$ term on the right hand side doesn't affect the frequency, notice that if we changed our origin so that $z = 0$ when $x = g/\omega^2$ -- that is, $z = x - g/\omega^2$ -- then the equation of motion for $z$ is $$\ddot{z} + \omega^2 z = 0$$

The equation of motion for the difference in position $d = x_2 - x_1$ of two identical masses attached to the ends of the spring in space is $$\begin{eqnarray} m \frac{d^2}{dt^2} \left(x_2 - x_1\right) &=& - k\left(x_2 - x_1\right) - k\left(x_1 - x_2\right) \\ &=& - 2 k\left(x_2 - x_1\right) \\ \end{eqnarray}$$ or $$\ddot{d} + \frac{2k}{m} d = \ddot{d} + \omega'^2 d = 0$$ where $2 \pi f' = \omega' = \sqrt{2k/m} = \sqrt{2} \omega = 2 \pi \left(\sqrt{2} f\right)$, so $$f' = \sqrt{2} f$$

Alternatively, think of the spring (with spring constant $k$) as two identical elementary springs (each with spring constant $k_0$) in series. It follows that $k=\frac{k_0}{2}$, and hence from the formula for the characteristic frequency, that $f=\frac{f_0}{\sqrt{2}}$.

1. In space without gravity, the midpoint will be at rest due to symmetry, so each side behaves as an elementary spring with frequency $f_0$.

2. Back on Earth, the spring will stretch in the gravitational field. Assuming that we stay within the linear regime of the spring, the spring constant will remain the same, so it will oscillate with the same characteristic frequency as in space, around the new equilibrium point. This fact can be seen as a result of a superposition principle in Newtonian mechanics.