Problem — Energy and amplitude

Problem

We want a simple harmonic oscillator ($m = 2\;\text{kg}$, $K = 800\;\text{N/m}$) to have total energy $E = 25\;\text{J}$. What amplitude, what maximum speed?

Solution

The total energy of a simple harmonic oscillator coincides with the potential energy at the point of maximum displacement, where all the energy is elastic:

$E = \tfrac{1}{2}K A^2 \;\Rightarrow\; A = \sqrt{\frac{2E}{K}} = \sqrt{\frac{50}{800}} = \sqrt{0{,}0625}\;\text{m}$

$\ev{A = 0{,}25\;\text{m}}$

When passing through equilibrium all the energy is kinetic, so:

$E = \tfrac{1}{2}m\, v_{\max}^2 \;\Rightarrow\; v_{\max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25}\;\text{m/s}$

$\ev{v_{\max} = 5\;\text{m/s}}$

Check with $\omega = \sqrt{K/m} = \sqrt{800/2} = 20\;\text{rad/s}$: indeed $v_{\max} = A\omega = 0{,}25 \cdot 20 = 5\;\text{m/s}$. ✓

Links

Topics: Oscillations and simple harmonic motion Concepts: Elastic potential energy · Kinetic energy · Conservation of mechanical energy Skills: Conservation of energy Objects: Spring