Question

A mass weighing 8 pounds is attached to a spring. When set in motion, the spring/mass system exhibits simple harmonic motion.

(a) Determine the equation of motion if the spring constant is$\mathbf{1}\mathbf{lb}\mathbf{/}\mathbf{ft}$and the mass is initially released from a point 6 inches below the equilibrium position with a downward velocity of$\frac{\mathbf{3}}{\mathbf{2}}\mathbf{ft}\mathbf{/}\mathbf{s}$.

(b) Express the equation of motion in the form given in (6).

(c) Express the equation of motion in the form given in $\left(6\text{'}\right)$.

Short answer

(a)So, the required equation is ${\mathit{x}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{3}}{\mathbf{4}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$.

(b)So, the required equation is $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\sqrt{\mathbf{13}}}{\mathbf{4}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2}\mathit{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{588}\mathbf{)}$.

(c) So, the required equation is $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\sqrt{\mathbf{13}}}{\mathbf{4}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9827}\mathbf{)}$.

Step-by-step solution

Definition of Gravity

Gravity is an invisible force that pulls objects toward each other.

(a)Evaluation

We are given a mass weighing$\mathit{m}\mathit{g}\mathbf{=}\mathbf{8}\mathbf{\text{lbs}}$that is attached to a spring and it set into motion.

Note that we need to rewrite the amount of the mass by dividing by gravity which is$\mathbf{32}\mathbf{\text{ft}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}$.

So, we have$\mathit{m}\mathbf{=}\frac{\mathbf{8}}{\mathbf{32}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}$slugs.

The spring has a spring constant of$\mathit{k}\mathbf{=}\mathbf{1}\mathbf{\text{lb}}\mathbf{/}\mathbf{\text{ft}}$.

We do not have to rewrite the spring constant.

We also have the following initial conditions:$\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\text{ft}}$and$\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{s}}$.

Note that we have to change the initial condition for the location of the mass into feet instead of inches.

General equation of Motion

The general equation of motion for a spring-mass system without any friction and no driving force is

$\mathit{m}{\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{k}\mathit{x}\mathbf{=}\mathbf{0}$

Using the information that$\mathit{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}$and$\mathit{k}\mathbf{=}\mathbf{1}$to set up the differential equation describing the equation of motion we get.

$\frac{\mathbf{1}}{\mathbf{4}}{\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathbf{=}\mathbf{0}$

Solving equation (1), we get the following auxiliary equation and roots.

localid="1668505367173" role="math" $\frac{\mathbf{1}}{\mathbf{4}}{\mathit{m}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$

$\frac{\mathbf{1}}{\mathbf{4}}{\mathit{m}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$

$\mathit{m}\mathbf{=}\mathbf{\pm }\sqrt{\mathbf{-}\mathbf{4}}$

$\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{2}\mathit{i}$

Use the roots in the equation

The roots of the auxiliary equation are complex so the solution of the differential equation takes on the following form.

${\mathit{x}}_{\mathbf{c}}\mathbf{=}{\mathit{e}}^{\mathbf{\alpha }\mathbf{t}}\left[c_{1}cos\left(\beta t\right)+c_{2}sin\left(\beta t\right)\right]$

Where the root of the auxiliary equation is of the form$\mathit{m}\mathbf{=}\mathit{\alpha }\mathbf{\pm }\mathit{\beta }\mathit{i}$.

So, the solution of the differential equation is

localid="1668505293024" ${\mathit{x}}_{\mathbf{c}}\mathbf{=}{\mathit{e}}^{\mathbf{0}}\mathbf{\left[}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}\mathbf{\right]}$

${\mathit{x}}_{\mathbf{c}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$

Use initial condition

Recall that we have the initial conditions of$\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\text{ft}}$and$\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{s}}$.

We can use the initial conditions to solve for constants${\mathit{c}}_{\mathbf{1}}$and${\mathit{c}}_{\mathbf{2}}$in equation (2).

Using the first initial condition of$\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$, we have

${\mathit{x}}_{\mathbf{c}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{2}\mathbf{·}\mathbf{0}\mathbf{)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2}\mathbf{·}\mathbf{0}\mathbf{)}$

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{·}\mathbf{1}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{·}\mathbf{0}$

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}$

Use second initial condition

Using the second initial condition of $\mathit{x}\mathbf{:}\left(0\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}$, we have to find the first derivative of equation

(2).

${\mathit{x}}_{\mathbf{c}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$

Now we plug in the initial condition to solve for${\mathit{c}}_{\mathbf{2}}$.

${\mathit{x}}_{\mathbf{c}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2}\mathbf{·}\mathbf{0}\mathbf{)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{2}\mathbf{·}\mathbf{0}\mathbf{)}$

$\frac{\mathbf{3}}{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$

$\frac{\mathbf{3}}{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{·}\mathbf{0}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

$\frac{\mathbf{3}}{\mathbf{4}}\mathbf{=}{\mathit{c}}_{\mathbf{2}}$

Using the constants we found, the equation of motion is ${\mathit{x}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{3}}{\mathbf{4}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$.

(b)Solve for constants

We want to rewrite equation (3) in the form of

$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{\omega }\mathit{t}\mathbf{+}\mathit{\varphi }\mathbf{)}$

We have the following

$\mathit{A}\mathbf{=}\sqrt{{\mathbf{c}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{2}}}$

$\mathit{\omega }\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

$\mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{c_1}{c_2}\right)$

So, we can find the three above constants.

$\mathit{A}\mathbf{=}\sqrt{{\left(\frac{1}{2}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{3}{4}\right)}^{\mathbf{2}}}$

$\mathbf{=}\frac{\sqrt{\mathbf{13}}}{\mathbf{4}}$

$\mathit{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{/}\mathbf{4}}}$

$\mathbf{=}\sqrt{\mathbf{4}}$

$\mathbf{=}\mathbf{2}$

$\mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{1/2}{3/4}\right)$

$\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{2}{3}\right)$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{588}$

Substitution

So, using the constants above and plugging into equation (4), we have

$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\sqrt{\mathbf{13}}}{\mathbf{4}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2}\mathit{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{588}\mathbf{)}$

$\mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{1/2}{3/4}\right)$

$\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{2}{3}\right)$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{588}$

(C) Rewrite the equation

We want to rewrite equation (3) in the form of

$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{\omega }\mathit{t}\mathbf{-}\mathit{\varphi }\mathbf{)}$

We have the following (note that the definition foris different)

$\mathit{A}\mathbf{=}\sqrt{{\mathbf{c}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{2}}}$

$\mathit{\omega }\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

$\mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{c_2}{c_1}\right)$

Solve for equation

So, the constants$\mathit{A}$and$\mathit{\omega }$are the same. We have to recalculate$\mathit{\varphi }$.

$\mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{3/4}{1/2}\right)$

$\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{3}{2}\right)$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9827}$

So, using the constants above and plugging into equation (5), we have$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\sqrt{\mathbf{13}}}{\mathbf{4}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9827}\mathbf{)}$