- #1
dRic2
Gold Member
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- Homework Statement
- A simple pendulum hangs from a fixed pulley. The other end of the string is in the hand of an observer who pulls up the string slowly, thus shortening the length of the pendulum with uniform velocity. Show that, neglecting friction, the amplitude of the oscillations increases in the following manner. The change of the total energy from the position ##\theta = 0## to the next position ##\theta = 0## is given by $$\Delta E = - \frac 1 2 \frac {\Delta l} l E$$ where ##E## is the energy constant of the undisturbed oscillations.
- Relevant Equations
- $$\Delta E = - \int_{t_1}^{t_2} \frac {\partial L} {\partial t} dt$$
Well, using the above equation it should be easy... but I can't solve it 
$$ L = \frac 1 2 m (\dot l^2 + l^2 \dot \theta ^ 2) - mgl(1- \cos\theta)$$
then I guess
$$\int_{t_1}^{t_2} \frac {\partial L}{\partial t} dt = L(t_2) - L(t_1)$$
*Note*: since the variation ##\frac {\partial L}{\partial t}## is considered wrt to the variable ##t## alone, in the infinitesimal time ##dt##, ##\theta## and ##\dot \theta## are kept constant. So, even when integrating ##\theta## and ##\dot \theta## are kept constant wrt to time. Am I right ? If I choose ##t_2## to be the instant when ##\theta## is again equal to zero and, of course, ##\theta(t_1) = 0## then:
$$\Delta E = -\frac 1 2 m (\dot l^2 + l^2(t_2) \dot \theta ^2) - \frac 1 2 m (\dot l^2 + l^2(t_1) \dot \theta ^2)$$
and ##l^2(t_2) = (l(t_1) + \Delta l)^2 \approx l^2(t_1) + 2l \Delta l##. So, finally,
$$\Delta E = - m (l(t_1)*\Delta l \dot \theta ^2)$$
But ##E_1 = \frac 1 2 m l^2(t_1)* \dot \theta ^2 = E## is the energy constant of the undisturbed oscillations. So:
$$\Delta E = - 2 \frac {\Delta l} l E $$
which is wrong.
$$ L = \frac 1 2 m (\dot l^2 + l^2 \dot \theta ^ 2) - mgl(1- \cos\theta)$$
then I guess
$$\int_{t_1}^{t_2} \frac {\partial L}{\partial t} dt = L(t_2) - L(t_1)$$
*Note*: since the variation ##\frac {\partial L}{\partial t}## is considered wrt to the variable ##t## alone, in the infinitesimal time ##dt##, ##\theta## and ##\dot \theta## are kept constant. So, even when integrating ##\theta## and ##\dot \theta## are kept constant wrt to time. Am I right ? If I choose ##t_2## to be the instant when ##\theta## is again equal to zero and, of course, ##\theta(t_1) = 0## then:
$$\Delta E = -\frac 1 2 m (\dot l^2 + l^2(t_2) \dot \theta ^2) - \frac 1 2 m (\dot l^2 + l^2(t_1) \dot \theta ^2)$$
and ##l^2(t_2) = (l(t_1) + \Delta l)^2 \approx l^2(t_1) + 2l \Delta l##. So, finally,
$$\Delta E = - m (l(t_1)*\Delta l \dot \theta ^2)$$
But ##E_1 = \frac 1 2 m l^2(t_1)* \dot \theta ^2 = E## is the energy constant of the undisturbed oscillations. So:
$$\Delta E = - 2 \frac {\Delta l} l E $$
which is wrong.
