- #1
CanIExplore
- 99
- 0
Hey all,
In the classical harmonic oscillator the force is given by Hooke's Law,
F = -kx which gives us the potential energy function V(x)=(1/2)kx^2. Now I understand that the first derivative at the point of equilibrium must be zero since the slope at the point of equilibrium is zero. But what is the meaning of the second derivative at that point? Is it how fast the slope is changing? My professor says that it has to be positive or equal to zero but that for simple harmonic motion it is always positive. I don't see how you can have the first derivative of a function evaluated at some point be zero but the second derivative at the same point be nonzero.
In the classical harmonic oscillator the force is given by Hooke's Law,
F = -kx which gives us the potential energy function V(x)=(1/2)kx^2. Now I understand that the first derivative at the point of equilibrium must be zero since the slope at the point of equilibrium is zero. But what is the meaning of the second derivative at that point? Is it how fast the slope is changing? My professor says that it has to be positive or equal to zero but that for simple harmonic motion it is always positive. I don't see how you can have the first derivative of a function evaluated at some point be zero but the second derivative at the same point be nonzero.