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One end of link chain is fixed to a support at \(z=0\). The free end of an, open, link-chain   of length $L$,  and weight $\mu$ per unit length, is released from rest at $x=0$. Determine the tension $T$ in the chain at its support at $A$ in terms of $x$. also determine the energy loss, $\Delta E$ during the entire motion $x=0$ to $x=2L$.

Source::IIT/K

A two stage rocket is fired as follows in deep space ( no gravity ).
Stage 1:
Mass of the frame+ engine = $M_1$
Mass of the fuel + oxidizer =$m_1$
gas velocity relative to the rocket = $v_r$
Stage 2:
After burning $M_1$ the second stage is ignited with
Mass of the frame+ engine = $M_2$
Mass of the fuel + oxidizer =$m_2$
gas velocity relative to the rocket = $v_r$
Show  that the velocity achieved is $$v_f= - v_r \ln \frac{(M_1+M_2+m_1+m_2)(M_2+m_2)}{(M_1+m_2+m_2)M_2} $$

   Source::HSM 70

 Water drop falling through saturated atmosphere.
A spherical  water droplet falls, without friction and under influence of gravity, through an atmosphere   saturated with water vapor. Let its initial radius ($t=0$)  be $c$, its initial velocity, $v_0$. As a result of condensation the  water drop experiences continuous increase in mass proportional to its surface; Show that its radius increases then increases linearly with time. Integrate the differential equation of motion by introducing $r$ instead of $t$ as an independent variable. Show that for $c=0$, the velocity increases linearly with time.

 Source::Sommerfeld

Without assuming that the Hamiltonian is independent of time, start from the requirement, \[ \begin{equation}
  _H{\langle \psi t \vert X_H(t)}_H\vert \psi \rangle_H = _{S\hspace{-1pt}}\langle\vert {\psi t}{X_S}\vert \psi\rangle_S \end{equation}\]  and directly derive the equation of motion \[ \begin{equation} \boxed{\frac{d X_H}{d t} = \frac{\partial X_H }{\partial t}+  \frac{1}{i\hbar}\big[X_H, H\big]_-.} \end{equation}\] for the general case when the Hamiltonian depends on time.

(a) For harmonic oscillator \[H=\frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2
\hat{x}^2\] set up and solve up the equations of motion for position and
momentum operators in the Heisenberg picture. Hence show that  \[\begin{eqnarray}    \hat{x}(t) = \tilde{x} \cos\omega t +
                   \frac{1}{m\omega} \tilde{p} \sin\omega t \\ \hat{p}(t) = \tilde{p} \cos\omega t - m\omega \tilde{x} \sin\omega t          \end{eqnarray}\]
where \(\tilde{x},\tilde{p}\) in the right hand sides of the above  equations are the  Schr\"{o}dinger picture operators.        
(b) Calculate the commutators   \[\begin{eqnarray*} [\hat{x}(t), \hat{x}(t^\prime))],\qquad  [\hat{x}(t),  \hat{p}(t^\prime))], \qquad  [\hat{p}(t), \hat{p}(t^\prime))],  \end{eqnarray*}\]   for \(t\ne t^\prime\). How do these commutators at unequal times
compare with equal time  commutators? Are \(\hat{x}(t),  \hat{x}(t^\prime)\) compatible observables?       
(c) Verify that at equal times the commutators reduce to canonical  commutation rule \([\hat{x}, \hat{p}] =i\hbar.\)

A particle is described by the Hamiltonian   $$ \hat{H} = \frac{\vec{p}\,^2}{2m} + V(\vec{x}) $$ Find the equations of motion of a particle in the Heisenberg picture. Solve the equations  of motion for the position and momentum $\hat{x}(t)$ and $\hat{p}(t)$. Find $x(t), p(t)$, if initial ($t=0$) conditions are given. Suppose the particle is in initial state $\psi$, what is the equation  satisfied by the expectation value of $x$ and $p$ at a later time $t$?  What does one expect classically?

A wire is bent in the shape of a parabola  \( x^2=4ay\).   A bead slides on the wire which is kept in a vertical plane  with \(y\)- pointing upwards.   How many generalized coordinates are needed? Obtain the Lagrangian and derive the Euler Lagrange equations of motion.

Let $C$ and $C^\prime$ be two neighbouring paths in configuration space
$$ C: q_k=q_k(t) \qquad    C^\prime : q_k = q_k(t) + \delta q_k(t) \qquad  t_1 < t < t_2 $$        
which do not have the same end points. Show that the difference in action along the two paths receives contribution only from the end points if and only if Euler Lagrange equations of motion are obeyed. In other words, show that the action  integral depends on the differences $\delta q_1 $ and $\delta q_2$ at the end points $t_1$ and $t_2$ only and not on values of $\delta q(t)$  for other values of $t$ between $t_1$ and $t_2$.

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