- Thread starter
- #1
Hi dwsmith!$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?
The book has a typo then.Hi dwsmith!
The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:
$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$