# Tag Info

18

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 ...

15

Assuming for a moment an infinitely hard and smooth surface, let's look at the energy of the ball. When the ball is dropped from a height $h$, initial potential energy is $mgh$. You would expect it to accelerate to a velocity $v=\sqrt{2gh}$. However, during the fall, it will experience drag from the air. This will cause the dissipation of some of the energy ...

12

When the ball hits the floor, its center of mass needs to be decelerated. It is done by deforming the ball (and possibly the floor if it is soft enough compared to the ball). In vacuum, if the ball and floor were perfectly elastic, they would then recover their previous shape following exactly the reverse dynamics as the deformation ones: in doing so, the ...

12

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 ...

10

@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 ...

10

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 ...

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 ...

8

Yes, if you burn it. Neglecting the metal, a $25\ g$ wooden mousetrap at $15\ MJ/kg$ should yield about $375\ kJ$.

8

Let me guess: you take the spring as it is and hang your objects, right? Then measure the displacement. Try to do the following: hang any arbitrary object so that the string will stretch a bit from its initial state. Then add you 100g and 200g objects to the initial mass and measure the difference in spring's length. I will be surprised if you won't get ...

8

You are asking two questions really 1) How is PE actually stored in a steel spring at the atomic level? The explanation for this lies in quantum mechanics 2) Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape? Replying to 1) ...

7

Let $E$ denote a quantity that does not change over time (from the first principle). Consider a ball with mass $m$ dropped from a height $h$. As the ball drops, its speed changes due to the gravitational acceleration $g$, reaching a final value $v$ at impact. Thus, we can infer that the quantity $E$ depends on these 4 parameters: $$E(m,H,g,V)$$ where $H$ ...

7

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

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

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

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 ...

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

Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work.

5

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = ... 5 You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds. It needs to be moving because, if it were not, then there ... 5 The second solution is there to allow for arbitrary start and stop times. Using standard trig identities you can convert an arbitrary linear combination of$\sin$and$\cos$into a time-displaced sinusoidal function: $$A\sin(\omega t)+B\cos(\omega t)=R\cos(\omega(t-t_0)),$$ where$R=\sqrt{A^2+B^2}$and$\tan(\omega t_0)=A/B$. 5 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, ... 5 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 ... 4 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: ... 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 The two oscillations will not be independent as they will share frequency and phase. You can start from Newton's equations of motion. $$m_1 \frac{{\rm d}^2\,a(t)}{{\rm d}t^2} = -F(t)$$ $$\mbox{-}m_2 \frac{{\rm d}^2\,b(t)}{{\rm d}t^2} = F(t)$$ $$F(t) = k \;\left( a(t)+b(t) \right)$$ Assume simple harmonic motion$a(t) = A \sin(\omega\,t)$,$b(t) = B ...

4

To supplement the answer by Luboš Motl, I will come to this problem from a Material Science point of view. What you mean by the inherent property of the string is not the spring constant, in fact, it is Young's modulus $E$, which only depends on the properties of a material of a body but not it's shape. $$E = \frac{\text{tensile stress}}{\text{tensile ... 4 Yes, it is a torsion spring. It works by twisting the metal rod that makes up the body of the spring. The reason for coiling the spring is to fit a long length of metal rod into a short space. You need a long length of rod so that the torsion per unit length remains small. With a shorter length of rod you'd exceed the elastic limit and the rod would be ... 4 When a metal spring is stretched beyond it's elastic limit, the metal begins to undergo some plastic deformation. This is a permanent deformation of metal crystals caused by the creation and motion of crystal lattice dislocations. These processes are partially irreversible and some of the work performed to deform the spring is lost as heat. 4 No, it is not. Your system will go through the same point twice in every oscillation, once moving in each direction, and the friction force will be reversed in each pass, so your approach doesn't work. What you need to consider is the velocity, not the displacement, so$$ma=-kx - \mathrm{sign}(v) F_{\mathrm{fric}}.$$This is not all that helpful in ... 4 The Lagrangian of the system is,$$L = \frac{1}{2}m\dot{x_1}^2 - \frac{1}{2}k_1(l_1 - l_0+x_1)^2-\frac{1}{2}k_2(l_2-l_0+x_2-x_1)^2+mgx_1 Here, $x_1$ is the downward distance from the equilibrium position of the mass, $x_2$ is the downward distance from the midpoint of the driving oscillator, $k_1$ and $k_2$ are spring constants of the top and bottom ...

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