# Tag Info

14

Well, the sentence It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why? clearly isn't a valid argument to calculate the $k$ of the smaller springs. They're different springs than their large parent so they may have different values of an "inherent property": if a pizza is divided to 4 smaller pieces, the ...

9

@HDE's comment's experimental approach done. Answer between about 1 and 2 ounces = 1/4 to 1/2 Newtons (http://www.wolframalpha.com/input/?i=convert+2+ounces+to+newtons). Key travel 3mm, say, = 1.5 milliJoules = $3.59×10^{-7}$ dietary Calories (http://www.wolframalpha.com/input/?i=convert+0.5+newtons+times+3+millimeters+to+Calories). So, for every ...

9

Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ...

8

The potential energy only being defined up to a constant does not imply that potential energy differences only depend on differences in position. To see this mathematically, assume that a function $U$ has the property that $U(x_2)-U(x_1) = f(x_2-x_1)$ for some function $f$. Then if we take $x_2 = x+\Delta x$ and $x_1 = x$, and divide both sides by ...

6

The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept $\beta$, your friend takes into account that the true zero of the mass parameter $m$ may be shifted from what you think it is. So an affine model $T^2=Cm+\beta$ is ...

6

You could make an analogy between the pressure distribution of a sound wave and the mass density distribution of a realistic spring undergoing vibrations, but it wouldn't give you the explanation you're looking for. As a matter of fact, that would be more like explaining a sound wave in terms of springs, rather than what you're trying to do, i.e. explaining ...

6

Kind of. The negative sign indicates the direction of the force exerted by the spring on the mass. If you pull the mass to the right, the force from the spring is to the left. Since they go opposite directions, there is a minus sign. The problem states an external force exerted on the mass displaces it, presumably to a new equilibrium. The spring ...

6

You seem to be more interested in energy than force. It is not possible to say "the force exerted on the keys thus far is enough to push a car five miles, or is equivalent to 100 kg of TNT" because pushing a car over a certain distance and exploding a certain amount of TNT aren't examples of a set force. The TNT has a certain amount of energy. Pushing the ...

6

What an awesome question! By the way, as far as I know, the original video is here for those interested. One key to understanding this is the following fact from classical mechanics that is a version of Newton's second law for systems of particles: The net external force acting on a system of particles equals the total mass $M$ of the system times the ...

4

Well, the reflection of a wave at the end happens always. One can picture this by imagining the succesive atoms being pushed off the equilibrium position as the wave propagates. Since the endpoint is fixed, it has nowhere to be pushed but the few atoms near it (I am considering idealized linear chain for simplicity) that have already being perturbed will, ...

4

I haven't used a mousetrap for several decades, but as I recall the moving arm is about 5cm long, so the tip moves 0.05$\pi$ or about 0.16m. To get 17J of work the force at the tip of the arm would need to be 100N. I'm fairly sure the force isn't anything like that great. I remember being able to pull the arm back with one finger. I would guess the force is ...

4

As long as the total weight does not exceed the elastic limit, then yes. Why? For any given length element, the weight that it is supporting doubles, so the change in effective length of that segment also doubles as per Hooke's law. The limit is when the most strained element passes it's elastic limit. That will be the topmost bit which is holding the ...

4

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 ...

3

You know the basic spring equation, right? $F = xK$, where $K$ is the spring constant, in units of force per distance. You also know work (energy) is force times distance, right? So all you've got to do is integrate $xK dx$ from d1 to d2. (Hint, you can pull $K$ out of the integral.) You could do it on graph paper if you happened to know d1 and d2. ADDED: ...

3

The reason for this result is the sign in you damping term. For a damped harmonic oscillator you need to have a resistive force on the mass point at $x$. That means if $x=0$ is the equilibrium position the damping term will be proportional to the velocity with an negative constant $F_{\text{damp}} = -ax',a>0$. I.e. the total force on the mass point ...

3

and welcome to the site. I suppose that you have to do this as some kind of homework so I would not try to invent new things and just look what mathematical description(s) of such a system exists. The governing equations will have some constants in it and it would be already a nice thing to look if the given framework can explain the dynamics and if you can ...

3

It depends also on the shape of the object. If you assume the trampoline is circular, and the object is much smaller (like a point mass) then you can start developing the equations. You have to know the initial tension of the trampoline, and also assume the material non-elastic but supsended by perfect strings in a radial direction (with known stiffness). ...

3

No, it is not. Your system will go through the same point twice in every oscillation, once moving in each direction, and the friction force will be reversed in each pass, so your approach doesn't work. What you need to consider is the velocity, not the displacement, so $$ma=-kx - \mathrm{sign}(v) F_{\mathrm{fric}}.$$ This is not all that helpful in ...

3

Yes. Some of the elastic energy stored in the spring does work by moving lattice dislocations through the metal - this is the physical mechanism responsible for the plastic deformation of the metal spring - and is the reason the spring may be permanently deformed when unloaded, even when the grip position applied to the spring has remained fixed. Plastic ...

3

This question is actually one of the lab exercises I teach. For a spring-mass system, if the damping force is friction, then it is independent of velocity (verified experimentally). However, as mentioned in the comments, the damping force may not always be friction. For example, if the mass is a material like aluminium and it is oscillating over some ...

2

Maybe they do; but Fourier transforms have an inherent flaw which makes them less than useful for such cases. That flaw is that they are transforms of steady state conditions: The initial and final conditions of the system are assumed to be the same; and transients are not considered. The transform for which you seek is the Laplace transform. Laplace is a ...

2

The units you mean are probably kg*cm (sometimes written kg.cm in robotics). Your original specification of 24 kgcm is a torque and not a force. The difference in practice is that, as the units imply, your resulting force at a point a distance from the "pivot point" decreases by the distance. So 24 kg*cm means that it can hold 24 kg at 1 cm or 12 kg at 2 cm ...

2

Since @jalaxiou provided a solution for the static case looking at the balance of forces, I will try to provide a solution for the dynamic case from energetic considerations. Ǹotation and energies I will first restrict my self to a 1D trampoline, i.e. a rope with 2 springs of stiffness $k$. The length of the string + the springs without pretention is ...

2

Springs in series The masses' finite lengths and the rest lengths of the springs are not really relevant because they don't change and hence don't contribute to the energy. If the total length from the beginning of the first mass to the end of the last one is 10 meters, for example, and all the rest lengths of springs and sizes of masses add up to 3 ...

2

I'm assuming you want to solve such problems automatically on a computer. If you want to solve them individually by hand, the best method will obviously be different. Without the constraint that the objects can't pass through each other, the problem is an unconstrained quadratic optimization problem, which is solved simply by taking the gradient of that ...

2

Expanding on my comment. A spring, or anything that can be modeled by a string is any interaction that can be described by a potential that is proportional to $x^2$ the displacement. Why is this potential important? Qualitatively, it is the solutions to this potential (i.e. the motion of a particle that sits in this potential) are the solutions of a ...

2

For a given spring, $k$ is a constant, As long as you're talking about an ideal spring. In other words, the definition of the ideal spring is that it applies the force proportional to its deformation length (at both endings of course). I'm afraid both you and your professor are wrong. The correct formula should be: $k_{new} = k_{orig}*4$ To show that ...

2

Lets suppose that all N magnets are identical. Assume that they are quite far from each other so we can replace them with N magnetic dipoles with dipole moment $\overrightarrow{m}$. Also assume that the dipoles are placed along the x axis in such a way that they repel each other and all N moments $\overrightarrow{m}$ are parallel. The first step is to ...

2

First, the fact that your graph goes to negative infinity would indeed mean that the spring will break, or at least that it's going to stop behaving like a spring. Basically, consider that you have obtained this solution using a model of a spring in which the force it exerts is proportional to its displacement from equilibrium. Once you reach the point at ...

2

The spring "with mass" has two possible complications: it's affected by gravity, and it may have a non-negligible kinetic energy (inertia). You say the spring is positioned horizontally - hence the gravity effect is irrelevant. As long as you consider only static problems - the kinetic energy is irrelevant as well, and your spring behaves exactly as ...

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