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Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes1.
You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, ...
77
I think that most of the answers here are incorrect since it has nothing to do with decreasing resistance of rubber. In fact, the force required to stretch the balloon increases, not decreases while inflating. It's similar to stretching a string, ie. the reaction force is proportional to the increase in length of the string - this is why there is a point ...
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There are two separate concepts here:
the Young's modulus, which determines the force needed to stretch the material
the elastic limit, aka yield strain, which determines how far the material can be stretched
As you say, the term elastic tends to be used in a vague way that conflates these two properties. Generally a high Young's modulus means the material ...
50
Both the OP and John Rennie have well illustrated the imperfections in the usage of the word "elastic" in physics and how the word can create confusion between "stiffness" and a material's ability to brook strain.
But an important point to be made is that the one important field where one hears the vague statement that "steel is more elastic than rubber" is ...
43
The coils of the spring are touching one another and the spring is initially under self-compression so it takes a finite force to move all the coils away from one another and for the spring to behave as you expected.
That initial part of the source vs stretch curve is real (you had obtained data whilst undertaking an experiment) and hence should not be ...
34
Take a strip of balloon rubber and pull it. It will get harder the more you pull. So why is it that inflating the balloon gets easier (at least long before the breaking point)?
The balloon starts with very high curvature, so the air pressure is distorting each spot on its surface a lot relative to its 1cm neighbors for example. All the rubber's tension ...
30
There is no "systematic error" in the experimental results. I have done this experiment as well when teaching a high school physics class. The effect you are seeing is due to the fact that the spring you are using requires a small amount of force before it begins to stretch, while the "theoretical" spring that is described by Hooke's Law does not. This ...
23
When in doubt, use mathematics.
Imagine the balloon as a sphere (close enough for this answer) of initial radius $r_0$ and thickness $t$. Let's inflate it just a little bit from the uninflated state (to radius $r_0 + \Delta r$). Now we can take a look at what happens by taking a cut through the equator of the sphere. The total circumference at the equator ...
17
Like Dev said above, the material your typical round balloon is made from has a non-linear stress strain curve. When just starting to inflate it is fairly stiff, but then as it starts to blow up the stiffness goes down somewhat until it approaches its maximum size. We measured this in my undergraduate advanced lab class, and while I don't have the data ...
16
Is it that the wooden block vibrates with lesser frequency than the
metal block? If so, why is that?
'Yes', to the first question.
Metal is stiffer than wood and produces higher frequencies (higher pitch).
This follows from the wave equation (here in one dimension):
$$u_{tt}=\frac{E}{\rho}u_{xx}$$
$E$ is Young's Modulus and $\rho$ the material's ...
15
The volume of a balloon grows linear, while the surface (which you actually stretch) doesn't. So although you're blowing the same amount of air into a balloon, you don't stretch the surface as much as in the beginning.
13
UPDATE :
After looking again at the video, I agree that Floris' explanation seems to be correct and my explanation below is wrong. Slightly different frequencies of vibration in two perpendicular planes accounts more simply for a rotation which reverses one way then the other. Kinetic energy seems to decay constantly; it does not seem to be stored in an ...
13
The metal block has a relatively low level of internal damping, however the wooden block has a high level of internal damping: Much of the energy imparted to the wooden block is dissipated internally as heat and deformation, also the higher frequencies are damped more than the lower frequencies (it acts like a low pass filter).
So the wooden block will ...
11
A toy model of a solid
A very simple model of a solid is to imagine a bunch of molecules linked to their nearby neighbors by springs (you don't need to imagine a crystalline lattice, it can be amorphous).
The springs are the effective electromagnetic interactions between the molecules; they are strongly repulsive at close range and attractive at modest ...
answered Aug 18 '12 at 20:59
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It is much easier to explain with a diagram than it is with a chart or equations.
Imagine threading tiny weightless disks along your length of blue wool. Twist it and hold it stretched out horizontally between your fingers. Each disk along the length of the wool is rotated a different amount (in a vertical plane) by the twist in the wool - disks at one end ...
9
Let's draw a comparison with ceramics, which—just as metals are generally ductile—are generally brittle.
First, note that crystals (and metals and ceramics are both generally polycrystalline) can deform through dislocation motion. A dislocation is a line defect that carries plasticity through a crystal. The classic analogy is moving a rug by kicking a ...
9
Yes, we can derive Hooke's Law from more basic continuum conditions, provided that the material be stable and at equilibrium, so that the strain energy is smoothly minimized with respect to the distance between atoms. (This energy is sometimes called the pair potential and is modeled using functions such as the Lennard-Jones potential.)
Consider just a pair ...
9
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to ...
8
Collisions can be elastic or inelastic.
Elastic collisions are collisions where the incoming and outgoing kinetic energies are the same and only the angles change
The same holds true classically, example: billiard balls ; and in the elementary particle framework. example: An electron hitting an electron has a probability of scattering elastically.
...
7
If you collide two ideal billiard balls, then that would be what you would call a perfectly elastic collision.
If you have a large dense collection of billiard balls, and you slam a new one into the collection, then there are a whole lot of elastic collisions, transfering energy and momentum in many ways that you would be hard-pressed to calculate exactly.
...
7
Referring to your graph which is for a ductile material I suggest the following.
A is the limit of proportionality up to which the stress and strain are proportional to one another and when unloaded the material goes back to its original length.
B is the elastic limit.
With stresses below this the material behaves elastically ie when unloaded returns ...
7
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a ...
7
This called a "twist to writhe" transition. It is much studied in biology! See the Wikipedia article "DNA supercoil."
I think it is easier to explain with energy rather that by studying forces.
The basic idea is that elastic energy of a flexible cable (or fibre or DNA molecule) can be approximated by
$$
E=\int_\gamma \left\{\frac 12 \alpha ( \omega\cdot{...
6
Hooke's Law is frequently used to model multi-dimensional materials because the stress tensor is simple (linear). The full expression can be found on Wikipedia. The simplification for 2D is straight forward (drop any terms with a 3 in the subscript). Note that whether deformation in one dimension affects the others is a property of the material and shows up ...
6
When a metal spring is stretched beyond it's elastic limit, the metal begins to undergo some plastic deformation. This is a permanent deformation of metal crystals caused by the creation and motion of crystal lattice dislocations. These processes are partially irreversible and some of the work performed to deform the spring is lost as heat.
6
Let us first summarize what do we actually experience when inflating the balloon. For the very first bit of volume, we have to exert a lot of energy. Or alternatively, we have to apply a lot of pressure coming from our lungs because for the change of energy $\delta E$, change of volume $\delta V$ and extra pressure $\Delta P$ (that is the difference between ...
6
Here's some tennis racket physics from Rod Cross, including links to several Am. J. Phys articles (the physics educators' journal, thus excellent for learning from) and this excellent diagram:
There are at least three "sweet spots":
The node, at the center of the strings, is a point where the natural standing waves in a vibrating racket don't have any ...
6
Will a tennis ball go further if i hit it with the side of the racket?
No.
You want the racket to deform, not the ball. This means using the strings to elastically store energy and return it to the ball.
The Ball
The ball's deformation upon impact is undesirable because "a tennis ball is required by the rules of tennis to dissipate a fraction of ...
6
I am sorry, I think I have misunderstood your question in my answer below.
You are asking about how long it will take the silicone to return to its original shape. This behaviour is called relaxation. Follow up on Gert's suggestion of studying viscoelasticity.
As Sanya says in his comments, the behaviour of visco-elastic materials is complex and non-linear....
6
In addition to sammy gerbil's answer, some energy is lost due to hysteresis.
This explains why cured (aka 'vulcanised') rubbers, like silicone rubber, have restitution coefficients that are less than unity.
These materials also show what is known in the industry as 'permanent set': when compressed (or extended) for a while they may not fully regain their ...
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