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What matters here is how the value of $c$ compares to the value of $k$. Let us choose a $\zeta = \frac{c}{2\sqrt{mk}}$ One can show that when $\zeta =1$ the system is critically damped, and will not exhibited any oscillations and will return to the origin in the shortest possible time interval. When $\zeta > 1$ the system is over damped and will take ...

1

You are correct, they are different. $k$ is not a property of the material, its a property of the entire object. Imagine having a small amount of a fairly tight spring. It takes a lot of effort to extend it even a centimeter or two. Now without changing the material, connect a few hundred of the springs together. Extending it a centimeter now will take ...

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The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ...

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As CuriousOne wrote, the device will lack frequencies ( and power ) The general solution is to evaluate the efficiency of Time reversal signal processing. If the bell ringing is , each time it is processed , identical ( rarely true ) the aerology between the bell and you is each time identical you are able to record all the sound and to reproduce it ...

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Good question. Consider the following two scenarios. 1) Your friend fell out the window while trying to stop his refrigerator from falling out the window and you caught him. He is dangling out of the window and you are holding him by the ankle. He is holding the refrigerator in his hand. His arm which is supporting the fridge is essentially a spring. 2) ...

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Newton's second law says $F=ma=m\frac{dP_m^2}{dt^2}$. Substituting yields $$m\frac{dP_m^2}{dt^2} = k(P_a-P_m)$$ Since $P_a$ is moving at constant velocity, we can substite $vt$ for $P_a$. Doing this and rearranging, we find $$m\frac{dP_m^2}{dt^2} + kP_m = kvt$$ This is indeed a differential equation, a second order inhomogeneous linear ordinary ...

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Replace spring B with a massless rod. Do you expect spring A to extend more, less or the same? Why? You can treat the whole hanging system of the two masses and the middle spring as black box. As long the the total weight does not change, it does not matter how much spring B deflects (or not in case of a rod). It is going to deflect as much as it is needed ...

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Substitute $x$ by $A(e^{wt})$. Thus your equation becomes: $$A(w^2)(e^{wt})+A(\frac{b}{m})(w)(e^{wt})+(\frac{k}{m})(A)(e^{wt})=0$$ Simplifying: $$(w^2) + (\frac{b}{m})(w) + (\frac{k}{m})=0$$ Find out the roots . Here you understand that $(b^2)<=4km$ for real values of $w$. Let the roots be $w_1$ and $w_2$ Finally your solution to the ...

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You have a correct answer, but I'm not sure if you have it for the right reason. Hooke's law states that $$\vec{F}_{\text{spring}} = -k\vec{r}$$ where $\mathbf{r}$ is the displacement of the spring from its equilibrium length (which you have not provided in the problem). If you take the equilibrium length to be 0, the spring force then has a magnitude ...

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If $S$ is de distance from $C$ to de $y$ axis, then $S=|r|\cos(\theta)$ where $r$ is the position of $C$ with respect to the origin and $\theta$ is the angle between the spring and the $x$ axis. However if you want to write $F_y$ then you would have $F_y=-k|r|\sin(\theta)u_y$. So you were correct in thinking in the decomposition of $F$ as a function of ...

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