I have an equation, and I want to know the physical significance of each of the following terms:

$E(t) = E(0) + \int_0^t g(s)u'(s)ds -\gamma\int_0^t[u'(s)]^2ds$

where $g = mu'' + \gamma u' + ku$. I know that E(0) is the initial energy (at time t=0), but I'm not sure what the next two terms mean physically. I also know that integrating these terms gives the energies, but I want to know what the terms themselves mean. Thanks!

It looks like $g$ represents the driving force of your oscillations, $u$ is the position of the mass $m$, $\gamma$ is the drag coefficient, and $k$ is the spring constant. Primes denote time derivatives.

As for the integrals, notice how they are integrals over a time interval ($ds$ has units of time). Since the integrals themselves must have units of energy, it must be that the integrands have units of energy/time, which represents power.

You can determine how much power is being supplied by a force $ \mathbf{F}$ by the relation $$P=\mathbf{F}\cdot\mathbf{v}$$ where $\mathbf v$ is the velocity of the object that the force is acting on.

Therefore you can then see that the first integral represents the energy supplied by the driving force, and the second integral is the energy dissipated by the drag force.

Your equation is then just an application of the definition of power $P=\frac{dE}{dt}$:

$$\int_0^t\frac{dE(s)}{ds}ds=\int_0^tP(s)ds$$ $$E(t)-E(0)=\int_0^t\left(\sum_i F_i(s)\cdot u'(s)\right)ds=\sum_i\int_0^t\left(F_i(s)\cdot u'(s)\right)ds$$

The spring force does not have an integral I am guessing because $E$ includes both kinetic and potential energy, so the spring force does not change the value of $E$.

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