Chapter 19.1, Problem 19.11P

To determine

(a)

The time period of the block to move 3 in upward.

Answer to Problem 19.11P

Time $t=0.0352s$

Explanation of Solution

Given information:

Mass of block $m=3lb$

Spring constant $k=2lb/in.$

Velocity $v=90in./s$

For simple harmonic motion:

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$

And, Natural frequency: ${\omega }_n=\sqrt{\frac{k}{m}}$

Where $k=2\times 12=24lb/ft$

$\begin{array}{l}{\omega }_n=\sqrt{\frac{24}{\left(\frac{3}{32.2}\right)}}\\ {\omega }_n=\sqrt{257.6}\\ {\omega }_n=16.04rad/s\end{array}$

When the box is in equilibrium,

$\begin{array}{l}x=0=x_m\sin \left(0+\varphi \right)\\ \varphi =0\end{array}$

$\begin{array}{l}x=\left(0.4673\right)\sin \left(16.05t\right)\\ x_m=\frac{7.5}{16.05}\\ x_m=0.4673ft\end{array}$

Now, $x=\left(0.4673\right)\sin \left(16.05t\right)\_\_\_\_\_\_(1)$

The time when $x=3in.=0.25ft$ is from equation (1):

$\begin{array}{l}0.25=\left(0.4673\right)\sin \left(16.05t\right)\\ \frac{0.25}{0.4673}=\sin \left(16.05t\right)\\ 0.534988=\sin \left(16.05t\right)\\ {\sin }^{-1}\left(0.534988\right)=16.05t\\ 32.3430976=16.05t\\ t=\frac{32.3430976}{16.05}\\ t=0.0352s\end{array}$

To determine

(b)

The velocity and acceleration of the block.

Answer to Problem 19.11P

Velocity $v=6.34ft/s$

Acceleration $a=-64.4ft/s^2\downarrow$

Explanation of Solution

Given information:

Mass of block $m=3lb$

Spring constant $k=2lb/in.$

Velocity $v=90in./s$

For simple harmonic motion:

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$

And, Natural frequency: ${\omega }_n=\sqrt{\frac{k}{m}}$

Where $k=2\times 12=24lb/ft$

$\begin{array}{l}{\omega }_n=\sqrt{\frac{24}{\left(\frac{3}{32.2}\right)}}\\ {\omega }_n=\sqrt{257.6}\\ {\omega }_n=16.04rad/s\end{array}$

When the box is in equilibrium,

$\begin{array}{l}x=0=x_m\sin \left(0+\varphi \right)\\ \varphi =0\end{array}$

$\begin{array}{l}x=\left(0.4673\right)\sin \left(16.05t\right)\\ x_m=\frac{7.5}{16.05}\\ x_m=0.4673ft\end{array}$

Now, $x=\left(0.4673\right)\sin \left(16.05t\right)\_\_\_\_\_\_(1)$

The time when $x=3in.=0.25ft$ is from equation (1):

$\begin{array}{l}0.25=\left(0.4673\right)\sin \left(16.05t\right)\\ \frac{0.25}{0.4673}=\sin \left(16.05t\right)\\ 0.534988=\sin \left(16.05t\right)\\ {\sin }^{-1}\left(0.534988\right)=16.05t\\ 32.3430976=16.05t\\ t=\frac{32.3430976}{16.05}\\ t=0.0352s\end{array}$

Then from the equation of simple harmonic motion:

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)or,$

$x=x_m\sin \left({\omega }_{n}t\right)\therefore \varphi =0$

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, $v=\frac{dx}{dt}$

$\begin{array}{l}v=\frac{d}{dt}\left(x_m\sin \left({\omega }_{n}t+\varphi \right)\right)\\ v={\omega }_n{x}_m\cos \left({\omega }_{n}t+\varphi \right)or,\\ v={\omega }_n{x}_m\cos \left({\omega }_{n}t\right)\end{array}$

By putting the above obtained value the velocity is calculated as;

$\begin{array}{l}v=\left(0.4673\right)\left(16.05\right)\cos \left(16.05\times 0.0352\right)\\ v=7.500165\cos \left(0.56496\times \frac{180}{\pi }\right)\\ v=7.500165\cos \left({32.386242}^{°}\right)\\ v=6.33ft/s\end{array}$

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

$\begin{array}{l}a=\frac{dv}{dt}\\ a=\frac{d}{dt}\left({\omega }_n{x}_m\cos \left({\omega }_{n}t+\varphi \right)\right)\\ a=-{\omega }_n^2{x}_m\sin \left({\omega }_{n}t+\varphi \right)or,\\ a=-{\omega }_n^2{x}_m\sin \left({\omega }_{n}t\right)\end{array}$

$\begin{array}{l}a=-\left(0.4673\right){\left(16.05\right)}^2\sin \left(16.05\times 0.0352\right)\\ a=120.377648\sin \left(0.56496\times \frac{180}{\pi }\right)\\ a=120.377648\sin \left({32.386242}^{°}\right)\\ a=-64.47ft/s^2\end{array}$