New answers tagged

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The "should have been proof" relies in the extensions of the springs being the same. If you put a=b that condition is not satisfied. Find a spot where the two extensions are the same and you will get "should have been proof" answer.


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Let's start by plotting the damped harmonic oscillation equation: $$x = Ae^{-\gamma t} \cos (\omega t)$$ In the above example $\gamma = 0.8$ and $\omega = 2.5\times\pi$. It should now be more clear that when $t = 0$, $x=A$ i.e. the initial displacement of the oscillator from it's equilibrium position. Here $A = 1~\mathrm{m}$. Let's assume down is the ...


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The answer is the opposite of what you would have if the springs were resistors.


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As we know, for a mechanical system being in equilibrium means having total force equal to zero. Let's look now at the second spring on a picture b): to be in equilibrium, force from the first spring $F_{k1}$ (action of the first spring on the second) should equalise external force $F_{out}$. Thus, $F_{out}=-F'_{k1}$. This, in turn, after applying Newton's ...


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Since the springs are connected in series, they will experience an equal amount of force as the tension is same inside both the springs in equilibrium. If the tension is non-zero, then some part of spring will accelerate and therefore equilibrium will be destroyed. In case of parallel springs, the force exerted by them will be the sum of the two forces as ...


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Dip773, you wrote the most general solution already. As stated by Floris, you can obtain a generalized formula ($k_{eff} = k_p\cdot\phi$, where $\phi = \frac{1+\sqrt{5}}{2}=1.61803...$ is the golden ratio) if all spring constants are equal ($k_n = k_p$ for each $n$). In order to obtain this, you also need the hypotesis that $n$ is large, and that adding one ...


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The $dx$ is actually a vector $d\vec{x}$ (because $\vec{x}$ is actually a vector). A vector's sign depends on your coordinate axes, e.g. if you pick right to be positive then left pointing vectors are negative. In general, the work done $W$ is given by $$ W = \int \vec{F} \cdot d\vec{x} = \int Fdx \cos(\theta) $$ where $\theta$ is the angle between ...


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It all depends on the direction you assign while using $dx$. It is generally a vector, so the sign will automatically be adjusted according the operation used and its direction relative to other operand vectors. In your case, you are calculating the work done by the spring. The force due to a spring is always directed to its mean position. With this in ...


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basic concept is to set up a coordinate system , Cartesian would be good here Here, I'm calculating work done BY the spring origin is the point where spring is in natural position (for us to use F = -Kx , otherwise, it would also contain some constant too, ex- if our spring's natural length is at (2,5) and spring only moves in along x axis direction, ...


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You might get a wrong sign due to $cos(0)= 1$ and $cos(\pi)= -1$. You either take $x$ as a cartesian coordinate and integrate over it or you use polar coordinates with some maximum elongation $x$ which is constant. In your question you are mixing up both. The conventional sign here is $-$, however it depends on your way of counting the work, ie work done to ...


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Even though im not completely sure, where your exact problem is, ill give it a try: Lets first think about a book lying directly on earth's surface: First of all theres a gravitational force $F_G=\frac{m_1\cdot m_2}{r^2}$ This force does not specify which object is "the pulling one" and which object is "the pulled one" - its just a force, acting between ...


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What would the force be if there was no spring? Each side of the spring feels the same force - so if you put a black box around the spring and only saw the string "going in" and a string "coming out" of the box, with the same tension on each, the force needed to move the box would be the same. This means your approach is correct.


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F(x) = kx, according to Hooke's Law. This means that your friend is incorrect, and you got the question correct. Note: the potential energy of the spring is (kx^2)/2, so your friend confused potential energy with force. A bit of dimensional analysis would greatly decrease the chance of mixing the units in this way.


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Let me assume(correct me if I am wrong) that the displacement is purely in the y-direction, starting from the free state of the spring. In such a case, the length of the spring can be calculated using the Pythagoras theorem(say l'). The change in length of the spring would now be l' - l( where l is the free length of the spring ). The spring force acting on ...


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Where in one dimension you would have $F=-kx$, you now have $\mathbf{F}=-k\mathbf{x}$ (assuming a simple Hookean spring) with $\mathbf{x}$ being the distance vector from the mass to the pivot of the spring.


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The force in the middle spring is $k(y_2-y_1)$ because it is lengthened when $y_1\lt 0$ or $y_2 \gt 0$. A positive sign on that force indicates that $m_1$ is pulled down and $m_2$ is pulled up. When $m_2$ is pulled down, $y_2 \gt 0$ and there is a downward force on $m_1$, so it will cause an increase in $\ddot {y_1}$. It will also cause an upward force on ...


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It's not that the final solution looks like that. Rather, you are looking for all the solutions of that form (normal modes) for two reasons: They are easy to find You can afterwards decompose any motion into a sum of normal modes. This comes from writing your equations of motion in the normal basis*. So first you work out the normal modes by assuming a ...


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Your first part was correct; but for the second part, you have to equate the energy of A plus the stored energy in the spring to the energy of B (because you start with no energy in the spring, and all the energy as kinetic energy in B). So the expression for the stored energy is $E_\mathrm{spring}=\frac12 k x^2 = \frac12 m_b v_b^2 - \frac12 m_a v_a^2$. ...


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Well I would say your first approach is absolutely correct and will result in right answer if the calculations are done correctly (I have not done the calculations) but your second approach is faulty. Problem with second approach In the second approach you have used the conservation of mechanical energy it seems. The problem is that conserving mechanical ...



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