Question

A mass weighing$\mathbf{64}$pounds stretches a spring$\mathbf{0}\mathbf{.}\mathbf{32}$foot. The mass is initially released from a point$\mathbf{8}$inches above the equilibrium position with a downward velocity of$\mathbf{5}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{s}}$.

(a) Find the equation of motion.

(b) What are the amplitude and period of motion?

(c) How many complete cycles will the mass have completed at the end of$\mathbf{3}\mathit{\pi }$seconds?

(d) At what time does the mass pass through the equilibrium position heading downward for the second time?

(e) At what times does the mass attain its extreme displacements on either side of the equilibrium position?

(f) What is the position of the mass at$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{\text{s}}$?

(g) What is the instantaneous velocity at$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{\text{s}}$?

(h) What is the acceleration at$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{\text{sA}}$?

(i) What is the instantaneous velocity at the times when the mass passes through the equilibrium position?

(j) At what times is the mass$\mathbf{5}$inches below the equilibrium position?

(k) At what times is the mass$\mathbf{5}$inches below the equilibrium position heading in the upward direction?

Short answer

(a)$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$

(b)$\mathit{T}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{5}}$

(c) 15

(d)$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7218}$

(e)$\mathit{t}\mathbf{=}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{\pi }}{\mathbf{20}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{09273}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}$

(f)$\mathit{x}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{597}\mathit{f}\mathit{t}$

(g)$\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{814}\mathit{f}\mathit{t}\mathbf{/}\mathit{s}\mathit{e}\mathit{c}$

(h)$\mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{59}\mathbf{.}\mathbf{702}$

(i)$\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\pm }\mathbf{8}\frac{\mathbf{1}\mathbf{\text{'}}}{\mathbf{3}}\mathbf{ }\mathit{f}\mathit{t}\mathbf{/}\mathit{s}\mathit{e}\mathit{c}$

(j)$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1451}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3545}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }$

(k)$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3545}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }$

Step-by-step solution

Definition

When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium.

Find mass & spring constant

Consider a mass weighing$\mathbf{64}$pounds stretches a spring$\mathbf{0}\mathbf{.}\mathbf{32}$foot.

From Hooke's law,

$$\begin{gathered} \mathbf{64}\mathbf{=}\mathit{k}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{32}\mathbf{)} \\ \mathit{k}\mathbf{=}\frac{\mathbf{64}}{\mathbf{0}\mathbf{.}\mathbf{32}} \\ \mathbf{=}\mathbf{200}\mathbf{\text{lb}}\mathbf{/}\mathbf{\text{ft}} \end{gathered}$$

Also$\mathit{W}\mathbf{=}\mathit{m}\mathit{g}$

The mass is:

$$\begin{gathered} \mathit{m}\mathbf{=}\frac{\mathbf{W}}{\mathbf{g}} \\ \mathbf{=}\frac{\mathbf{64}}{\mathbf{32}} \\ \mathbf{=}\mathbf{2}\mathbf{\text{slugs}} \end{gathered}$$

Find differential equation

Substitute the above values in the differential equation$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\left(\frac{k}{m}\right)\mathit{x}\mathbf{=}\mathbf{0}$.

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\left(\begin{array}{c}200\\ 2\end{array}\right)\mathit{x}\mathbf{=}\mathbf{0} \\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{100}\mathit{x}\mathbf{=}\mathbf{0} \end{gathered}$$

Give the mass is initially released from a point 8 inches above the equilibrium position.

So,$\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{8}}{\mathbf{12}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}$. Since$\mathbf{1}\mathit{f}\mathit{o}\mathit{o}\mathit{t}\mathbf{=}\mathbf{12}$inches.

The downward velocity is$\mathbf{5}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{sec}}$i.e;${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$.

Thus, the initial value problem is$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{100}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$.

Find general solution(a)

The equation of motion.

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{100}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}$

The auxiliary equation is${\mathit{m}}^{\mathbf{2}}\mathbf{+}\mathbf{100}\mathbf{=}\mathbf{0}$.

$$\begin{gathered} {\mathit{m}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{100} \\ \mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{10}\mathit{i} \end{gathered}$$

Thus, the general solution of the differential equation is$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{10}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{10}\mathit{t}$.

Apply the initial condition $\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$.

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

Apply initial condition

Differentiate$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}$with respect to$\mathit{t}$.

$$\begin{gathered} {\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{10}{\mathit{c}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}\mathbf{10}\mathit{t}\mathbf{+}\mathbf{10}{\mathit{c}}_{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{10}\mathit{t} \\ {\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{10}{\mathit{c}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}\mathbf{10}\mathit{t}\mathbf{+}\mathbf{10}{\mathit{c}}_{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{10}\mathit{t} \\ \mathbf{=}\mathbf{-}\mathbf{10}{\mathit{c}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}\mathbf{0}\mathbf{+}\mathbf{10}{\mathit{c}}_{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{0} \\ \mathbf{5}\mathbf{=}\mathbf{10}{\mathit{c}}_{\mathbf{2}} \\ {\mathit{c}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Thus, the solution is$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{10}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{10}\mathit{t}$.

Find amplitude & phase angle

Write the equation$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}$in the form$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{w}\mathit{t}\mathbf{+}\mathit{\varphi }\mathbf{)}$where amplitude$\mathit{A}\mathbf{=}\sqrt{{\mathbf{c}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{2}}}$and phase angle defined by$\mathit{A}\mathbf{=}\sqrt{{\mathbf{c}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{2}}}$.

Here${\mathit{c}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}{\mathit{c}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

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

The angle is,

$$\begin{gathered} \mathit{t}\mathit{a}\mathit{n}\mathit{\varphi }\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{{\mathbf{c}}_{\mathbf{2}}} \\ \mathbf{=}\frac{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}}{\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}} \\ \mathit{\varphi }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(-\frac{4}{3}\right) \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273} \end{gathered}$$

Substitute the above values in$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{w}\mathit{t}\mathbf{+}\mathit{\varphi }\mathbf{)}$. Thus, the equation of the motion is.$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$

Find the amplitude and period of the motion.(b)

From part (a),

The amplitude$\mathit{A}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathbf{\text{ft}}$.

The period of the motion is,

$$\begin{gathered} \mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\omega }} \\ \mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{10}} \\ \mathbf{=}\frac{\mathbf{\pi }}{\mathbf{5}} \end{gathered}$$

Find the number of complete cycles(c)

The number of complete cycles mass have completed at the end of$\mathbf{3}\mathit{\pi }$seconds.

The mass can complete one cycle in$\frac{\mathbf{\pi }}{\mathbf{5}}$seconds.

$$\begin{gathered} \mathbf{3}\mathit{\pi }\mathbf{=}\frac{\mathbf{\pi }\mathbf{k}}{\mathbf{5}} \\ \mathit{k}\mathbf{=}\mathbf{15} \end{gathered}$$

Thus, number of complete cycles will the mass have completed at the end of $\mathbf{3}\mathit{\pi }$seconds are 15.

Find time(d)

If$\mathit{x}\mathbf{=}\mathbf{0}$and the weight is moving downward for the second time, then

$\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathbf{0}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathbf{0}$

Since$\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{\pi }\mathbf{=}\mathbf{0}$where$\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{\dots }$

The mass pass through the equilibrium position for the second time i.e;$\mathit{n}\mathbf{=}\mathbf{2}$ .$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{\pi } \\ \mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{=}\mathbf{2}\mathit{\pi } \\ \mathbf{10}\mathit{t}\mathbf{=}\mathbf{2}\mathit{\pi }\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273} \\ \mathit{t}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{218}}{\mathbf{10}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7218} \end{gathered}$$

Find displacement(e)

Differentiate of$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$with respect to.

${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$

Extreme displacements are attained when${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

$$\begin{gathered} \frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathbf{0}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathbf{=}\mathbf{0}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathbf{=}\mathit{c}\mathit{o}\mathit{s}\left(n\pi +\frac{\pi }{2}\right)\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273} \\ \mathbf{=}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}\mathbf{10}\mathit{t} \\ \mathbf{=}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}\mathit{t} \\ \mathbf{=}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{\pi }}{\mathbf{20}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{09273}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.} \end{gathered}$$

Thus, the extreme displacement on either side of the equilibrium position is$\mathit{t}\mathbf{=}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{\pi }}{\mathbf{20}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{09273}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}$.

Position of mass(f)

The position of the mass at$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{\text{sec}}$is given by:

Substitute$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{\text{sec}}$ in$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$.$$\begin{gathered} \mathit{x}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{10}\mathbf{\right(}\mathbf{3}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{29}\mathbf{.}\mathbf{0727} \\ \mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{597}\mathbf{\text{ft}} \end{gathered}$$

Instantaneous velocity(g)

The instantaneous velocity at$\mathit{t}\mathbf{=}\mathbf{3}$is given by:

Substitute$\mathit{t}\mathbf{=}\mathbf{3}$in${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$.$$\begin{gathered} {\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{10}\mathbf{\right(}\mathbf{3}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathbf{29}\mathbf{.}\mathbf{0727} \\ \mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{814}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{sec}} \end{gathered}$$

Find the acceleration at.(h)

For acceleration differentiate${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$with respect to$\mathit{t}$.$$\begin{gathered} {\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{25}}{\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{\left(}\mathbf{10}\mathbf{\right)} \\ \mathbf{=}\mathbf{-}\frac{\mathbf{250}}{\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \end{gathered}$$

Substitute$\mathit{t}\mathbf{=}\mathbf{3}$in ${\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$.

$$\begin{gathered} {\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{250}}{\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{10}\mathbf{\right(}\mathbf{3}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathbf{=}\mathbf{-}\frac{\mathbf{250}}{\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathbf{29}\mathbf{.}\mathbf{0727} \\ \mathbf{=}\mathbf{59}\mathbf{.}\mathbf{702} \end{gathered}$$

Instantaneous velocity(i)

The mass passes through the equilibrium position, $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$,

$$\begin{gathered} \frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathbf{0} \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{\pi } \\ \mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{=}\mathit{n}\mathit{\pi } \\ \mathbf{10}\mathit{t}\mathbf{=}\mathit{n}\mathit{\pi }\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273} \\ \mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{(}\mathit{n}\mathit{\pi }\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.} \end{gathered}$$

Substitute$\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{(}\mathit{n}\mathit{\pi }\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$in${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}$.

The instantaneous velocity at the time is,

$$\begin{gathered} {\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\left(\frac{3}{10}(0.9273+n\pi )-0.9273\right) \\ \mathbf{=}\mathbf{\pm }\mathbf{8}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{sec}} \end{gathered}$$

Time for displacement below equilibrium position(j)

We have$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{5}$inches.

role="math" localid="1663970661171" $\mathit{x}\mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}\mathbf{\text{ft}}$

$$\begin{gathered} \mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \frac{\mathbf{5}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{5}}{\mathbf{6}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)} \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{10}\mathit{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{9273}\mathbf{)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\left(\pm \frac{\pi }{6}+2n\pi \right)\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots } \\ \mathbf{10}\mathit{t}\mathbf{=}\mathbf{\pm }\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{+}\mathbf{2}\mathit{n}\mathit{\pi }\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9273} \\ \mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\left(\pm \frac{\pi }{6}+2n\pi +0.9273\right) \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1451}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3545}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots } \end{gathered}$$,

Find time(k)

The mass 5 inches below the equilibrium position heading in the upward direction.

That is${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{<}\mathbf{0}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}$

$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3545}\mathbf{+}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{5}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }$is satisfies above two conditions.