Find the constant of motion

Classical Mechanics Level 2

Consider a particle with mass $m=3~\text{g}$ that can move on the x-axis and interacts with a wave moving to right with speed $u=1000~ \text{m}/\text{s}$ . The interaction energy between the particle and the wave is given by $V(x-u t)$ where $V(x)$ is an unknown function. Because the interaction energy is time-dependent the usual total energy of the particle $E(t)= \frac{m \dot{x}^2}{2}+V(x-u t),$ is not conserved. That is, $\frac{dE(t)}{dt}\neq 0$ . Nonetheless, one can show that for certain $\beta$ the quantity $I(t)=\frac{m\dot{x}^2}{2}+V(x-ut)-\beta \dot{x}$ is conserved, i.e., $\frac{dI(t)}{dt}=0$ . Find $\beta$ in kg m/s .

The answer is 3.

2 solutions

Jiahai Feng
Jan 4, 2014

The extra term appears because $V$ is a function of $x - ut$ , instead of just a function of $x$ .

$\frac{dI(t)}{dt} = \frac{dE(t)}{dt} - \beta \ddot{x} = 0$

$\beta \ddot{x} = m\ddot{x}\dot{x} + \frac{d V(x-ut)}{d(x-ut)} \cdot \frac{d (x-ut)}{dt} = m\ddot{x}\dot{x} + \frac{d V(x-ut)}{d(x-ut)}\cdot(\dot{x} - u)$

And since $\frac{d V(x-ut)}{d(x-ut)} = -m\ddot{x}$ ,

We have $\beta\ddot{x} = um\ddot{x} \Rightarrow \beta = um = 0.003 \times 1000 = 3$

Mandar Sohoni
Feb 19, 2014

This is actually pretty straightforward, just remember that d[V(x-ut)]/d(x-ut) = Force, i.e = ma. Then just use the chain rule correctly and you should end up with B=mu => B=3 kgm/s

Find the constant of motion

The answer is 3.

2 solutions

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