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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 ...


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 ...


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

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

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 ...


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 $H$ ...


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

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 ...


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 ...


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 $\tan(\omega t_0)=A/B$.


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

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

I made a typing mistake yesterday. The correct period of oscillation should be $$ T=2\pi\frac{(2Mm)^{0.5}}{{(m+2M)^{0.5}}{(2k)^{0.5}}}. $$ It can be derived by applying NLMs & constraint relations & conservation of energy to finding time period in oscillatory motion. Now, my answer is completely correct. I have also varified it in a numerical of ...


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

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

In a sense yes, if you're very careful about what you're holding constant. Stating that a variable is proportional to another variable implies that all other relevant quantities are being held constant. For example, there's a simple relation $d=vt$ that describes the distance $d$ something travels in a time $t$ when traveling at speed $v$. One might say ...


4

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 ...


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

To supplement the answer by LuboŇ° 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 is defined to be the same for some material (look for a heuristic derivation bellow): $$ E = \frac{\text{tensile ...


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

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.


3

Given a spring with spring constant $k$ whose extension is in the direction $x$, the magnitude of the force that the spring exerts is given by $$ |F| = k|L-l| $$ where $L$ is its length and $l$ is its equilibrium length. Now, imagine that the two masses are at positions $x_1$ and $x_2$ with $x_2>x_1$, then the length of the spring is given by $L = x_2 ...


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 ...



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