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- Thread starter dwsmith
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- Mar 5, 2012

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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 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:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$

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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:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$