According to the answer key, the kinetic as well as potential energy is maximum in the mean position. I am unable to understand the reason behind this. Please explain.
120828
In wave motion of a string both kinetic energy and potential energy are minimum at $y=y_\text{max}$ then why does the string come down again? As everything in nature tries to attain the lowest energy possible, what brings that string element back to its original position?
25781
For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$ \frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$  From this, it follows there's also a corresponding scalar quantity conserved:$$\frac{d}{dt}\sum\limits_i^n m_i(v^2_{i1} + v^2_{i2} + v^2_{i3}) = \frac{d}{dt}\sum\limits_i^n m_iv^2_i =0,\quad j=1,2,3$$  So why is there a need to put 1/2 in front of this conserved scalar quantity, kinetic energy?
35643
I have read the explanation for this in several textbooks, but I am struggling to understand it via Archimedes' principle. If someone can clarify with a diagram or something so I can understand or a clear equation explanation that would be great.
1199
Consider a rocket in deep space with no external forces. Using the formula for linear kinetic energy $$\text{KE} = mv^2/2$$ we find that adding $100\ \text{m/s}$ while initially travelling at $1000\ \text{m/s}$ will add a great deal more energy to the ship than adding $100 \ \text{m/s}$ while initially at rest: $$(1100^2 - 1000^2) \frac{m}{2} \gg (100^2) \frac{m}{2}.$$ In both cases, the $\Delta v$ is the same, and is dependent on the mass of fuel used, hence the same mass and number of molecules is used in the combustion process to obtain this $\Delta v$. So I'd wager the same quantity of chemical energy is converted to kinetic energy, yet I'm left with this seemingly unexplained $200,000\ \text{J/kg}$ more energy, and I'm clueless as to where it could have come from.
67162
The following quote comes from:      Consider an ice skater that has just gone into a spin with arms   stretched out.  If it helps, the skater is wearing lead bracelets on   each wrist.  As the skater spins, the angular momentum and angular   kinetic energy are constant;  friction with the air and with the ice   can be eliminated.  The skater must exert an inward centripetal force   on the bracelets to keep them rotating in the same circle, but this   force does no work, since the bracelets are not changing their radius   of rotation.      Then the skater pulls in his arms and the bracelets, closer to his   axis of rotation.      Several interrelated things happen:           since there is no external torque, the angular momentum of the system stays the same;   Since the various masses of the skater are all moving towards the axis of rotation, the total moment of inertia of the system   decreases;     Combining 1 and 2, the angular velocity of the system increases;  if the moment is halved, the angular velocity doubles   Combining 2 and 3, with some quantitative treatment, the angular kinetic energy increases;  if the moment is halved and the   angular velocity is doubled, the kinetic energy doubles;   The skater does work;  his arms are now exerting an inward force through a distance as his arms draw the bracelets inward;  the work done can be rigorously shown to be identical to the increase in   kinetic energy...      However, it only covers a case when work is done to pull arms inwards. It is understandable that the work would be done in such case.  But what is happening in the opposite case when the skater has their arms inwards and lets them go outwards? S/he doesn't do any work but kinetic energy (and total energy) is lost. Why is it lost?  E.g.  I = 2kg * (1m)^2 w = 3.14/s L = 2kg * m^2 * 3.14/s Ek = 9.869  After shifting 2kg by 0.1m outwards: I = 2kg * (1.1m)^2 = 2.42 kg * m^2 L = 2 * 3.14 = 2.42 * w w = 2.5963 Ek = 8.1566   So, where did 1.7124 J go?  The quoted answer describes the easy case and omits the hard one.
31867
I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding).   If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:  Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?     Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment Oh! It's because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:  **What goes on in the resistor that is literally removing electrical energy **. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks!
45177
I'm having trouble understanding how a  (not the  that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I don't see how this alignment gives rise to "attraction" or force.
70630
Basically, which is the correct formula for thermal energy, and is this the same as kinetic energy? My notes are pretty conflicting on this topic, and I'm getting pretty confused.
8169
Yesterday, while studying a simple question, a rather strange doubt hit my mind:     Consider a ball of mass $m$ moving with velocity $v$ that collides with a wall in a perfectly elastic collision and rebounds back with the same velocity. Taking the direction of initial motion of the ball as positive, the change in momentum of the ball is $-2mv$ and the wall is clearly at rest.   So, in order to obey the law of conservation of momentum, the wall must gain a momentum of $2mv$.  Now, this itself confused me a lot. I then applied the law of conservation of kinetic energy to this situation, and clearly the magnitudes of the initial and final velocities of the ball are equal, so kinetic energy of the ball is conserved and hence the final kinetic energy of the wall must be zero in order to conserve energy.  Now, this contradicts with the law of conservation of momentum, which says that the wall must gain momentum of $2mv$.  One more point: If we are talking about a wall then it is an important fact, which must be considered, that the wall is fixed on earth and so the mass of the wall is equivalent to that of earth, say $M$. If the law of conservation of momentum is applied now then we get velocity of the wall, say $V=2mv/M$ which is no doubt an extremely small value. But, the law of conservation of energy strictly says that the velocity of the wall must be exactly zero, not even the smallest value is allowed (theoretically speaking).   Please clarify this and correct me anywhere if I'm wrong. I have tried my best to make it clear what I want to ask :).
10214
I'm having trouble understanding the motivation between wanting to transition between the two descriptions.  I figure that switching from  is useful if we want to recover a particle's trajectory maybe? Is there any other reason to do so? and then what are we measuring by switching from Lagrangian to Eulerian?
78570
My teacher told  me that pressure variation is maximum at mean position in a sound wave. Though by proof of pressure wave and displacement wave, I am convinced but in understanding the physical meaning I am facing a problem as I have a counter argument. According to me pressure variation should be maximum there at extreme position because  pressure is maximum due to the fact that density is maximum as all particles come closest to each other due to compression.Hence by this argument the claim that pressure variation is maximum at mean position contradicts my understanding. please help
11832
I suspect this has something to do with thermodynamics and the  and I'm interested in a mathematical derivation of this result.
112778
Let's start with kinetic energy (from )     The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed $v$ is $\frac{1}{2}mv^2$.   Let's say you &amp; your bike have mass of 100kgs, then your kinetic energy at 10m/s would be  $$ E_a = 1/2 \times 100 \times 10^2 = 5000J = 5kJ$$  If you apply another 5kJ of energy, you don't get to 20m/s though, you only get to:  $$ E_b = 10000J =1/2 \times 100 \times V_b^2$$    $$\implies V_b = \sqrt{10000 / (1/2 \times 100)}         = √200         = 14.14m/s$$  Let's say you and a buddy are both coasting along at 10m/s though, from their perspective you've just burned 5kJ but only accelerated 4.1m/s, even though you seemed stationary.  Imagine you and your mate are in space drifting along together, at an unknown speed. Your mate fires his burners and accelerates away from you. There's a big screen on his ship showing how many joules of energy he just burned, and you can measure his resulting relative velocity just fine.  The question is, Will 5kJ of energy always produce 10m/s of relative velocity,,assuming 100kg spaceships?  If 5kJ always produces 10m/s, Why does the second 5kJ only produce 4.1m/s? What is going on here?
120491
I am currently studying for an exam in Quantum Mechanics and came across a solution to a problem I have trouble with understanding.  The Problem:  A Particle sits in an infinite potential well described by  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  We know that the energies are given by $E_n = \dfrac{n^2 \pi^2 \hbar^2}{2 m L^2}$ and $\Psi(x) = A_n \sin(n \pi x /L)$.  At time $t_0$ the potential well is suddenly doubled in size, such that the potential is now  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq 2L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  So the energies are now given by $\tilde{E}_n = \dfrac{n^2 \pi^2 \hbar^2}{2 \cdot 4 m L^2}$ and $\tilde{\Psi}(x) = \tilde{A}_n \sin(n \pi x /2L)$.   If the particle is in the ground state of the potential well before the change, what is the probability to find the particle in the ground state of the new potential after the change?   This is absolutely clear to me. We find a non vanishing probability as a result. But now it gets tricky:   What is the expectation value of the energy of the particle directly after the change? How does the expectation value of the energy evolve in time?   The solution suggests that the expectation value of the energy does not evolve in time, which is clear to me, since the Hamiltonian is time independent and thus energy is conserved. But it also suggests that the expectation value does not change after we double the width of the potential wall which I understand from the argument of energy conservation but not in terms of quantum mechanics. If the probability that the particle is in the state $\tilde{\Psi}$ does not vanish the particle could have the energy $\tilde{E}_n$ which is lower than $E_n$ and this would mean that the expectation value of energy could change (with a given probability).  What am I missing here, where is my mistake? Any help is appreciated!
99963
Can someone explain me what this equation means? I read it from page 152 on .
21767
Just like classical potential, it's stated that equilibrium is obtained when the corresponding thermodynamics potential reaches the minimum. Explicitly, according to , in particular:  When the entropy (S ) and &quot;external parameters&quot; (e.g. volume) of a closed system are held constant, the internal energy (U ) decreases and reaches a minimum value at equilibrium. When the temperature (T ) and external parameters of a closed system are held constant, the Helmholtz free energy (F ) decreases and reaches a minimum value at equilibrium. When the pressure (p) and external parameters of a closed system are held constant, the enthalpy (H ) decreases and reaches a minimum value at equilibrium. When the temperature (T ), pressure (p) and external parameters of a closed system are held constant, the Gibbs free energy (G ) decreases and reaches a minimum value at equilibrium.  There is something confuse me here. Take the first claim for example. If we have the formula: $dU=TdS-pdV$ then when $S$ and $V$ are held constant, is it not clear that $dU=0$ for whatever changes occurring? Furthermore, if there are $two$ fixed parameter, is it not true that the state of system is fixed? (assuming there is a state formula $f(T,p,V)=0$). It seems like I'm really misunderstanding some concepts of the thermodynamics potential. Can someone give me a clear explanation for my confusion?
59488
This is quoted from A.P.French's Vibrations &amp; Waves:  If the driving force is of low frequency  relative to the natural frequency, we would expect the particle to move essentially with the driving force. This is equivalent in saying that $m\dfrac{d^2x}{dt^2}$ in $$m\dfrac{d^2x}{dt^2} + kx = F_0\cos\omega t$$ plays small role compared to the term $kx$. The amplitude is controlled by spring constant. On the other hand, at frequencies very large compared to the natural frequency, the opposite situation holds. The $kx$ becomes small compared to $m\dfrac{d^2x}{dt^2}$. In this case, we expect a relatively smaller amplitude of oscillation than the above case.  $\bullet$ Why does $kx$ become prominent in the first case &amp; not in the second case? $\bullet$ What is the physical reason behind that the first case has greater amplitude &amp; the second one smaller?