91Ó°ÊÓ

A simple harmonic oscillator's amount of energy is a motion constant and is given by

$E=\frac{1}{2}k{A}^2$

The total kinetic energy for a mass m object bouncing at the tip of a force nearly constant spring are given by

$K=\frac{1}{2}m{v}^2$

$U=\frac{1}{2}k{x}^2$

In basic harmonic motion, an oscillator's angular frequency is given by

$U=\frac{1}{2}k{x}^2$

Because the oscillator's energy is retained, the amount of energy is constant and equal to the sum of the oscillator's kinetic and potential energies:

Substituting we get,

$\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$

$k{A}^2=m{v}^2+k{x}^2$

Solving

$A=\sqrt{\frac{m{v}^2+k{x}^2}{k}}$

$A=\sqrt{\frac{(0.1\mathrm{kg})(0.2\mathrm{m}/\mathrm{s}{)}^2+(2.5\mathrm{N}/\mathrm{m}\left)\right(-0.05\mathrm{m}{)}^2}{2.5\mathrm{N}/\mathrm{m}}}$

$A=0.064m.$

The oscillator angular frequency is

$\mathrm{Ó\neg }=\sqrt{\frac{k}{m}}$

Substituting we get,

$\mathrm{Ó\neg }=\sqrt{\frac{2.5\mathrm{N}/\mathrm{m}}{0.1\mathrm{kg}}}$

role="math" localid="1651382625317" $\mathrm{Ó\neg }=5^{s-1}$

The maximum acceleration block is,

$a_{\max }={\mathrm{Ó\neg }}^{2}A$

$a_{\max }={\left(5{\mathrm{s}}^{-1}\right)}^2(0.064\mathrm{m})$

$a_{\max }=1.60\mathrm{m}/{\mathrm{s}}^2.$

Because when component is at its second lowest deviation from the equilibrium point, the acceleration has a peak size and a positive value; at zero the amplitude. So

$x=-A=-0.064\mathrm{m}.$

The equation as

$m{v}^2=k{A}^2-k{x}^2$

role="math" localid="1651382919403" $m{v}^2=k({\mathrm{A}}^2-{\mathrm{x}}^2)$

Solving for $v$,we get

$v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

Substituting,

$v=\sqrt{\frac{(2.5\mathrm{N}/\mathrm{m})\left[(0.064\mathrm{m}{)}^2-(0.03\mathrm{m}{)}^2\right]}{0.1\mathrm{kg}}}$

$v=0.283\mathrm{m}/\mathrm{s}.$