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

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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:

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

  • $\begingroup$ 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 $k$? $\endgroup$ – Farcher Oct 26 '17 at 22:06

Imagine throwing a ball straight up. It travels upwards while negatively accelerating due to the force of gravity, stops at its highest point (called the "apogee"), then comes back down. At the apogee, is the acceleration zero?

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.


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