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Some hints/guesses: probable driving forces are: the flag instability buckling, from the horizontal bending of the band caused by wind drag (try to bend a steel rule in its plane, it will buckle because it's much easier to bend orthogonally to its plane) the wavelength is linked to the intensity of the restoring force when the band is deformed. E.g. in ...


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There is a treatment of lowering a string through a Rindler horizon here, (which contains a brief discussion on the extent to which the approximation is representative).


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Any time you have an equal and opposite force acting on two bodies, the effect is proportional to $\frac{1}{m}$ for each one. The combined effect is proportional to $$\text{(effect)} = \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \text{(action)}$$ When inverted to find which action has a desired effect we get the reduced mass $$ \text{(action)} = \left( ...


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When you state that the two masses are 'kept at rest' I assume you mean that the pulley is prevented from turning. The load on the pulley is then (M+m)g. [If the pulley is not prevented from turning, then to keep the masses at rest we would have to lift the heavier mass M until it became, effectively, the same weight as the smaller mass. The total load ...


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When the blocks are kept at rest , for the vertical equilibrium the pulley has to exert the force equal to the weight of two blocks. But when the heavier block starts moving down the mechanical advantage of pulley with respect to the heavier block increases and hence the weight on the pulley decreases.


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When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...


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The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations. The tension is ...


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Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc


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The net force on the string is not F. You are pulling the string forward with force F but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to F. If the masses of the string and block are m and M then for the whole system (string plus block) F=(M+m)a. The net force on the string is F '=ma=mF/(M+m). ...


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You wrote it yourself, $F=ma$, so with zero mass there is zero force needed for a finite acceleration. A finite force gives an infinite acceleration. ... Lol, I just reviewed your answer about the electrostatic field being normal on the surface of a conductor. That's exactly the same, here! With a non-zero force the acceleration would be infinite ($F/m$ ...


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There are two very good answers here, don't you want to accept one? I also think that this is a good conceptual question, this is not clear to many. Both answers explained the constancy of the strain under the nececcary conditions. I will add one point which is not quite clear from the others: Your statement about the pulley is only true if (1) there is ...


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The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...


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My proposed explanation is in the thickness. We wish the headphone wires thin and shoelaces thick. The former favors fixing sharp bends by entangling them randomly, the latter favors straightening them. If we tolerated 5 mm thick nylon yarn around our headphones, they would certainly tangle much less.


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The important thing is the difference in tension between the ropes on either side of the pulley. So you will have $(T_2-T_1) R = I_{\text {pulley}}\alpha$ where $T_1$ and $T_2$ are the tensions on either side of the pulley, $R$ is the radius of the pulley, $I_{\text {pulley}}$ the moment of inertia of the pulley and $\alpha$ the angular acceleration of the ...


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The rotation will not necessarily be parallel to the ground. The general motion will be a combination of rotation in a horizontal plane (the conical pendulum) and oscillation in a vertical plane (the simple pendulum). If the support (pivot) is a fixed point, the motion you get depends on the starting conditions. If you launch the mass horizontally at the ...


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The definition given by mit is perfectly correct. I don't see any problem with it.


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Actually to get an exact expression for a hanging wire with a lumped mass somewhere in the middle is extremely hard (maybe impossible). Another thing to consider is the allowable sag of the rope. If the rope can sag a lot then the tension can be low, whereas if low sag is required the tension has to be really high. In addition as the tension increases the ...



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