# Considering the SMH of a particle in an idealized $1D$ potential energy well

I think I need a bit of guidance trying to understand the following exercise:

Here is my thought process:

• From my knowledge, for (a), the motion should be of the form $U = \frac{1}{2} kx^2$, so a quadratic form.
• For (b), however, I have a bit of confusion. Intuitively, when a mass undergoing simple harmonic motion has maximum potential energy, it is at max displacement in its motion $x_0$, and for a very small increment of time is motionless. Therefore, acceleration is $0$. However, I will try to attempt to show this mathematically.

$$U = \frac{1}{2} kx^2$$ $$U_{max} = \frac{1}{2} k x_{max}^2$$

For $x$ to be at its highest value, $x = x_0$.

$$\therefore U_{max} = \frac{1}{2} k x_0^2$$ $$\ x(t) = x_0 \cos(\omega_0t)$$ $$a = -x_0 \ \omega_0^2 \cos(\omega_0t)$$

I don't know how to arrive to my answer from here. It should be zero, intuitively, but I don't how my thought process is helping me here.

• For (c), I need to understand (b) to tie it in altogether, but I do have this:

$$U_{max} = \frac{1}{2}kx_0^2$$

• This question is about the basic properties of a body undergoing sham. Think about the definition of shm in terms of acceleration and displacement. How is the natural frequency related to the mass and your constant $k$? – Farcher Oct 26 '17 at 22:06

It cannot be, since the ball begins to accelerate back downward. The same principle applies to the simple harmonic oscillator: at the ends of its trajectory (aptly named "turning points"), it must have a non-zero acceleration since the restoring force tries to move the mass back toward the equilibrium point. Remember, in a harmonic oscillator $F =-kx$ (since $F = \frac{dU}{dx}$), so therefore force (and hence acceleration) is of greatest magnitude at the largest value of x.