What is the formula for max kinetic and max potential energy of a spring?
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2$\begingroup$ Writing "concept" in the title doesn't really change this from a "give me the formula" question. As always you would use the definition of work and because your context is springs you would use Hooke's law for a first approximation. $\endgroup$– dmckee --- ex-moderator kittenNov 20, 2012 at 14:26
2 Answers
I think you mean the energy of a particle attached to a spring. In that case the total energy is given by $$E=K+P=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}$$ Where $m$ is the mass of the particle, $v$ is the velocity of the particle, and $x$ is the distance of the particle from the origin. So, the maximum of potential energy $P$ is given when the kinetic energy $K$ is zero (i.e. when $v=0$, in the turning points of the particle). In that case $E=P_{max}$. So $$E=P_{max}=\frac{1}{2}kx_{max}^{2}$$ here $x_{max}$ is the maximum stretch of the spring (i.e. the turning point, $v=0$).
In the other hand, the maximum of kinetic energy is given when $P=0$ (it occurs when $x=0$), i.e. $E=K_{max}$, or $$E=K_{max}=\frac{1}{2}mv_{max}^{2}$$ where $v_{max}$ is the maximun velocity of the particle, it occurs when the particle passes through the origin (i.e. when $x=0$).
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$\begingroup$ Seems good but a better term for $x$ would be the displacement from the origin. $\endgroup$ Apr 16, 2014 at 8:09
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Well, what's the equation for the kinetic and potential energies of the spring in general? Based on that, for what values of displacement will they be maximized?