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Suppose we have a mass on a spring with a damping term. The equation of motion is given by:

$$m \ddot{x} = -kx - c\dot{x}$$

I believe solutions are damped oscillations of the form:

$$x = x_0 e^{-\alpha t}\cos{\omega t}$$

where $x_0$ is the initial extension of the mass.

I am interested in calculating the total energy taken out of the system by the damping term $F_{damp}=-c\dot{x}$:

$$E_{out}=\int_0^\infty F_{damp}(t)v(t) dt$$

$$E_{out}=\int_0^\infty -c \dot{x}\ \dot{x}\ dt$$

Would one expect this energy to always exactly balance the $1/2 k x_0^2$ that we put into the system initially?

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  • $\begingroup$ You solution is not complete - you need also an initial condition for $\dot{x}(0)$, and there are two linearly independent solutions for a second-order differential equation. See, e.g. Wikipedia on linear damping. Also, yes, one would of course expect total energy to be conserved - if you actually calculate $E_\text{out}$, you will find that it indeed is. $\endgroup$
    – ACuriousMind
    Feb 15 '15 at 15:33
  • $\begingroup$ Your balance should be $\frac{1}{2} k x_0^2 + E_{out} = 0$ with the solution of $k=m (\alpha^2+\omega^2)$ $\endgroup$ Feb 15 '15 at 17:06
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The damping introduces a dissipative element to the system, that is, energy is leaving the spring-mass system with time. As such, the energy is not conserved.

The maximum energy for the system occurs at its initial configuration; the spring is stretched to some extent $A$ (initial amplitude). The initial energy for the spring is thus purely of a potential nature. Now, recall that the potential energy of this system can be written \begin{equation*} E_{\text{total}}=U(A)=\frac{1}{2}kA^{2} \end{equation*} where $A$ is the initial amplitude. You have yourself supplied the time-evolution of the system, $x=x(t)$, which incorporates the amplitude. However, due to this damping the amplitude is time-dependant and we have \begin{equation*} A(t)=x_{0}e^{-\alpha t} \end{equation*} thus resulting in \begin{equation*} E_{\text{total}}(t)=\frac{1}{2}kx^{2}_{0}e^{-2\alpha t} \end{equation*} which is an explicit expression of the energy in the spring-mass system at times $t$. With this, we can write the energy lost to the environment (due to the damping friction) as \begin{equation*} E_{\text{out}}(t)\equiv E_{\text{total}}(t=0) - E_{\text{total}}(t) = \frac{1}{2}kx^{2}_{0}(1-e^{-2\alpha t}) \end{equation*}

Edit: What you mean by ''balance the initial energy $\frac{1}{2}kx^{2}_{0}$'' is to me unclear. But I believe that you may find the answer to that last question of yours by considering my answer.

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Would one expect this energy to equal the $1/2 kx_0^2$ that we put into the system initially?

Yes, it must.

Energy conservation applies between any two points in time. If your point one is the initially stretched out spring before it is let go with energy $U_{el}=\frac{1}{2}kx_0^2$ and your point two is the situation of no energy stored in the spring anymore $U_{el}=0$, then:

$$\sum E_{total,1}=\sum E_{total,2} \Rightarrow \\ U_{el,1}=U_{el,2}+Q \Rightarrow \\ \frac{1}{2}kx_0^2=Q_{damping}$$

Where $Q$ is heat or work done causing damping.

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  • $\begingroup$ Ok - so I just need to do the maths to verify that fact. $\endgroup$ Feb 15 '15 at 15:34
  • $\begingroup$ @JohnEastmond No need to verify. This is true. (Unless you have other forces interacting dragging out or adding energy.) Energy conservation is a law of nature. when you calculate the left side, you have the value of all energy dissipated, which I called $Q_{damping}$ here. $\endgroup$
    – Steeven
    Feb 15 '15 at 15:40
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There is a simple way to check that your formula for $E_{out}$ is correct and you do not even need an explicit solution to the differential equation:

If the formula is correct, then it must hold due to the conservation of the total energy that $E_{kin} + E_{pot} - E_{out} = \text{const}$. So we plug in your formula and check via differentiation: \begin{align*} \frac{d}{dt}\left( \frac12 m \dot{x}(t)^2 + \frac{1}{2}kx(t)^2 + c\int_0^t \dot{x}(t)^2\right) &= m {\dot x}(t){\ddot x}(t) + kx(t){\dot x}(t)+c{\dot x}(t)^{2} \\ &= {\dot x}(t)\bigl(-kx(t)-c{\dot x}(t)\bigr) + kx(t){\dot x}(t)+c{\dot x}(t)^{2}\\ &=0, \end{align*} where we have used the differential equation itself to arrive from the first to the second line.

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