An object attached to a spring undergoes simple harmonic motion modeled by the differential equation $\displaystyle {m}\frac{{{d}^{{2}}{x}}}{{{\left.{d}{t}\right.}^{{2}}}}+{k}{x}={0}$ where $\displaystyle {x}{\left({t}\right)}$ is the displacement of the mass (relative to equilibrium) at time $\displaystyle {t}$, $\displaystyle {m}$ is the mass of the object, and $\displaystyle {k}$ is the spring constant. A mass of $\displaystyle {5}$ kilograms stretches the spring $\displaystyle {0.65}$ meters.

Use this information to find the spring constant. (Use $\displaystyle {g}={9.8}$ meters/second²)

$\displaystyle {k}=$ Preview Question 6 Part 1 of 3  

The previous mass is detached from the spring and a mass of $\displaystyle {17}$ kilograms is attached. This mass is displaced $\displaystyle {0.3}$ meters below equilibrium and then launched with an initial velocity of $\displaystyle {2}$ meters/second. Write the equation of motion in the form $\displaystyle {x}{\left({t}\right)}={c}_{{1}}{\cos{{\left(\omega{t}\right)}}}+{c}_{{2}}{\sin{{\left(\omega{t}\right)}}}$. Do not leave unknown constants in your equation.

$\displaystyle {x}{\left({t}\right)}=$ Preview Question 6 Part 2 of 3  

Rewrite the equation of motion in the form $\displaystyle {x}{\left({t}\right)}={A}{\sin{{\left(\omega{t}+\phi\right)}}}$. Do not leave unknown constants in your equation.

$\displaystyle {x}{\left({t}\right)}=$ Preview Question 6 Part 3 of 3