Energy in Simple Harmonic Motion Video Lessons

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# Problem: A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the block has velocity -4.00 m/s and displacement +0.200 m.Find (a) the amplitude and (b) the phase angle Write an equation for the position as a function of time. Assume x(t) in meters and t in seconds.

###### FREE Expert Solution

(a)

Mechanical energy of a SHM oscilator:

$\overline{)\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}{{\mathbf{A}}}^{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{{\mathbf{v}}}_{{\mathbf{x}}}}^{{\mathbf{2}}}}$

At t = 0, x = 0.200 m and v = -4.00 m/s

$\begin{array}{rcl}\mathbf{A}& \mathbf{=}& \sqrt{\frac{{\mathbf{kx}}^{\mathbf{2}}\mathbf{+}\mathbf{m}{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}}{\mathbf{k}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{\left(}\mathbf{300}\mathbf{\right)}{\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{\left(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\right)}{\mathbf{\left(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{300}}}\end{array}$

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###### Problem Details

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the block has velocity -4.00 m/s and displacement +0.200 m.

Find (a) the amplitude and (b) the phase angle

Write an equation for the position as a function of time. Assume x(t) in meters and t in seconds.