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The phenomenon you're looking at is creep, in which a metal held at a constant strain will gradually relax and so reduce the stress. This will reduce the stored strain energy. It has nothing to do with fatigue, which is the result of repeated cyclic loading, and it has nothing to do with the homogeneity of the stress field. The Wikipedia article ...


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I cannot comment on the full design because the question is lacking on details, but I can explain more about the situation where you hit a mass with a ball and a spring reacts to it (like the sketch shown) Consider the friction force as $f=\mu m g$ and the equations of motion $$m \frac{{\rm d}^2x}{{\rm d}t^2} + k x \pm f = 0$$ The sign of $f$ depends on ...


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Did you see section 4 of the calculator "warnings"? "Your index is too large" (index = ratio of diameter of spring to outer diameter. You have a ratio of 40 - a very thin wire wound on a very large diameter. And since length inside hook needs to be decreased by the diameter in order to get the length of the "actual spring", you basically have "almost no ...


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To help you with section a: Your approach in writing the differential equation for the two positions (http://oi62.tinypic.com/dfx6yq.jpg) is correct. Since we know gravity won't affect the resonant frequency, I'll ignore gravity when writing the equations. $$\ddot{y_1} = -(k/m_1)(y_1-y_2)$$ $$\ddot{y_2} = -(k/m_2)(y_2-y_1) $$ Notice that the spring force ...


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The job of the tension is to keep the masses at a fixed length apart. The weight (and weight difference) plays no direct role in determining the tension. Consider two connected weights in free fall. The tension would be zero because both are moving at the same rate (same acceleration). So the job of the tension is to keep the rope length fixed, which ...


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There's no inconsistency because the system isn't in static equilibrium. The net force on the larger mass is $2mg-T$ and it will accelerate downwards. The net force on the smaller mass is $T-mg$ and it will accelerate upwards. If the string is inextensible, the two accelerations have equal magnitude and you can solve for $T$. BTW, you shouldn't put ...


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There are two points to note: Firstly, the tension T1 and T2 is equal only when the pulley and the rope are massless. If the pulley has mass=> it will have considerable inertia momentum. Due to its angular acceleration, its torque will be different from 0 , which implies that T1 and T2 is not equal. Also, if the rope is not massless, the tension will also ...


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First of all, you say how then can tensions forces in this photo be equal if their sources are of different weight This shows a fundamental misunderstanding. The two weights are not the "sources" of the tension. The tension results from the interaction between the whole rope and both masses. Somewhat expanding on the good answer from @Eeko, you ...


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I think you're misunderstanding what it means when by "the pulling force equals the tension force". Imagine pulling on a rope with the other end not fixed to anything. Even if you pull with a large force, the tension in the rope will be zero, since the whole thing is accelerating due to that force. So what's important to understand from this is that tension ...


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I believe the discrepancy is due to a incorrect stiffness of 18.15 N/m in the measurement based on displacement. If I assume correctly you define displacement as the how much the mass displaces when moved to the center (by hand?) to when it moved down and only held by the springs. If so, you have to take into account that in the center point the springs ...


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You made a mistake in your spring calculation. From the diagram, the two springs are in series, not parallel. These add in inverse. $$k_{effective} = (\frac{1}{k_1}+\frac{1}{k_2})^{-1} = k/2 $$ You predicted the effective spring constant to be around 8.87N/m, so you would expect the individual spring constants to be around 17.7N/m. This is much closer to ...


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You seem to have a hidden assumption in your work, that the springs are uncompressed before you displace them and take your measurement. It is easily possible that there is something in the mechanical setup of your device which keeps the springs compressed when your geophone is in its "resting" state.


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Hooke's law describes the force a spring exerts when its length is changed. Of course it makes intuitive sense that when you try to change the length of a spring (by pressing it, for example) the spring exerts a force on your hand, and resists the compression. For Hooke, this suggested that the force exerted by a spring is always in the opposite direction ...


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Compare Hooke's law, $F=-kx$, with Newton's 2nd law, $F=ma$: $$ m\frac{d^2x}{dt^2}=-kx\tag{1} $$ where $[k]=\rm N/m=kg/s^2$ and $[m]=\rm kg$. (1) can also be written as, $$ \frac{d^2x}{dt^2}=-\frac{k}{m}x\tag{2} $$ So $[k/m]=\rm 1/s^2$. Why not make up a new variable, call it $\omega$, with units of $\rm1/s$, such that (2) becomes, $$ x''=-\omega^2x $$ i.e., ...



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