Chapter 19.2, Problem 19.37P

The 9-kg uniform rod AB is attached to springs at A and B, each of constant 850 N/m, which can act in both tension and compression. If end A of the rod is depressed slightly and released, determine (a) the frequency of vibration, (b) the amplitude of the angular motion of the rod, knowing that the maximum velocity of point A is 1.1 mm/s.

Fig. P19.37

To determine

The frequency $\left(f_n\right)$ of vibration.

Answer to Problem 19.37P

The frequency $\left(f_n\right)$ of vibration is $\underset{\_}{3.65\text{Hz}}$.

Explanation of Solution

Given information:

The mass (m) of the uniform rod AB is 9 kg.

The spring constant of spring (k) A and B is $850\text{N/m}$.

The length (l) is 600mm.

Calculation:

Show the free body diagram of the rod AB as in Figure (1):

Write the expression for the force at A $\left(F_A\right)$:

$F_A=k\left[0.6l\theta +{\left({\delta }_{st}\right)}_A\right]$

Here, k is the spring constant and ${\left({\delta }_{st}\right)}_A$ is static deflection at A.

Write the expression for the force at B $\left(F_B\right)$:

$F_B=k\left[0.4l\theta +{\left({\delta }_{st}\right)}_B\right]$

Here, ${\left({\delta }_{st}\right)}_B$ is static deflection at B.

The external forces in the system are spring force and force due to the mass of the rod. The effective force in the system is $m\overline{a}$ and the couple in the system is $\overline{I}\ddot{\theta }$.

Taking moment about C,

${\displaystyle \sum M_C}=-0.6l\left(F_A\right)+0.1lmg-0.4l\left(F_B\right)$

Substitute $k\left[0.6l\theta +{\left({\delta }_{st}\right)}_A\right]$ for $F_A$ and $k\left[0.4l\theta +{\left({\delta }_{st}\right)}_B\right]$ for $F_B$.

${\displaystyle \sum M_C}=-0.6l\left(k\left[0.6l\theta +{\left({\delta }_{st}\right)}_A\right]\right)+0.1lmg-0.4l\left(k\left[0.4l\theta +{\left({\delta }_{st}\right)}_B\right]\right)$                  (1)

Take moment about C in the system for effective forces.

${\displaystyle \sum {\left(M_C\right)}_{eff}=\overline{I}\alpha +0.1lm{a}_t}$

Equate the moment about C in the system for external and effective forces

$\begin{array}{l}\sum M_C={\left(\sum M_A\right)}_{eff}\\ -0.6l\left(k\left[0.6l\theta +{\left({\delta }_{st}\right)}_A\right]\right)+0.1lmg-0.4l\left(k\left[0.4l\theta +{\left({\delta }_{st}\right)}_B\right]\right)=\overline{I}\alpha +0.1lm{a}_t\end{array}$            (2)

At the equilibrium position $\theta$ is 0.

Substitute 0 for $\theta$ in equation (1).

$\begin{array}{l}-0.6l\left(k\left[0.6l\left(0\right)+{\left({\delta }_{st}\right)}_A\right]\right)+0.1lmg-0.4l\left(k\left[0.4l\left(0\right)+{\left({\delta }_{st}\right)}_B\right]\right)=0\\ -0.6lk{\left({\delta }_{st}\right)}_A+0.1lmg-0.4lk{\left({\delta }_{st}\right)}_B=0\end{array}$ (3)

Substitute equation (3) in equation (2),

$\begin{array}{l}-0.6l\left(k\left[0.6l\theta +{\left({\delta }_{st}\right)}_A\right]\right)+0.1lmg-0.4l\left(k\left[0.4l\theta +{\left({\delta }_{st}\right)}_B\right]\right)=\overline{I}\alpha +0.1lm{a}_t\\ {\left(-0.6l\right)}^{2}kl\theta -0.6lk{\left({\delta }_{st}\right)}_A+0.1lmg-{\left(0.4l\right)}^{2}kl\theta -0.4lk{\left({\delta }_{st}\right)}_B=\overline{I}\alpha +0.1lm{a}_t\\ {\left(-0.6l\right)}^{2}kl\theta -{\left(0.4l\right)}^{2}kl\theta =\overline{I}\alpha +0.1lm{a}_t\\ {\left(0.6l\right)}^{2}kl\theta +{\left(0.4l\right)}^{2}kl\theta =-\overline{I}\alpha +0.1lm{a}_t\\ \overline{I}\alpha +0.1m{a}_t+{\left(0.6l\right)}^{2}k\theta +{\left(0.4l\right)}^{2}k\theta =0\\ \overline{I}\alpha +0.1m{a}_t+0.52l^{2}k\theta =0\end{array}$ (4)

Write the expression for moment of inertia $\left(\overline{I}\right)$ of the rod:

$\overline{I}=\frac{1}{12}m{l}^2$

Write the expression for the tangential acceleration $\left(a_t\right)$:

$a_t=0.1l\alpha$

Calculate the natural circular frequency $\left({\omega }_n\right)$:

Substitute $\frac{m{l}^2}{12}$ for $\overline{I}$, $0.1l\alpha$ for $a_t$ and $\ddot{\theta }$ for $\alpha$ in equation (4).

$\begin{array}{c}\frac{m{l}^2}{12}\ddot{\theta }+0.1m\left(0.1l\ddot{\theta }\right)+0.52l^{2}k\theta =0\\ \left(\frac{m{l}^2}{12}+{\left(0.1l\right)}^{2}m\right)\ddot{\theta }+0.52l^{2}k\theta =0\end{array}$

Substitute 9 kg for m and 600 mm for l.

$\begin{array}{l}\left(\frac{m{l}^2}{12}+{\left(0.1l\right)}^{2}m\right)\ddot{\theta }+0.52l^{2}k\theta =0\\ 0.09333m{l}^2\ddot{\theta }+0.52l^{2}k\theta =0\end{array}$

Divided the above equation by $m{l}^2$.

$\begin{array}{l}\frac{0.09333m{l}^2\ddot{\theta }}{m{l}^2}+\frac{0.52l^{2}k\theta }{m{l}^2}=0\\ 0.09333\ddot{\theta }+\frac{0.52k\theta }{m}=0\end{array}$

Divide the above equation by 0.09333.

$\begin{array}{l}\frac{0.09333\ddot{\theta }}{0.09333}+\frac{\frac{0.52k\theta }{m}}{0.09333}=\frac{0}{0.09333}\\ \ddot{\theta }+\frac{5.5716k\theta }{m}=0\end{array}$

Substitute $850\text{N/m}$ for k and 9 kg for m.

$\begin{array}{l}\ddot{\theta }+\frac{5.5716\times 850}{9}\theta =0\\ \ddot{\theta }+526.2\theta =0\end{array}$ (5).

Compare the differential Equation (5) with the general differential equation of motion $\left(\ddot{x}+{\omega }_n^{2}x=0\right)$ and express the natural circular frequency of vibration:

$\begin{array}{c}{\omega }_n=\sqrt{526.2}\\ =22.939\text{rad/s}\end{array}$

Calculate the frequency $\left(f_n\right)$ using the relation:

$f_n=\frac{{\omega }_n}{2\pi }$

Substitute $22.939\text{rad/s}$ for ${\omega }_n$.

$\begin{array}{c}{f}_n=\frac{22.939}{2\pi }\\ =3.65\text{Hz}\end{array}$

Therefore, the frequency $\left(f_n\right)$ of vibration is $\underset{\_}{3.65\text{Hz}}$.

To determine

The amplitude $\left({\theta }_m\right)$ of the angular motion of the rod.

Answer to Problem 19.37P

The amplitude $\left({\theta }_m\right)$ of the angular motion of the rod is $\underset{\_}{0.0076°}$.

Explanation of Solution

Given information:

The mass (m) of the uniform rod AB is 9 kg.

The spring constant of spring (k) A and B is $850\text{N/m}$.

The length (l) is 600mm.

The maximum velocity ${\left({\dot{x}}_A\right)}_m$ of point A is $1.1\text{mm/s}$.

Calculation:

Write the expression for angular deflection:

$\theta ={\theta }_m\sin \left({\omega }_{n}t+\varphi \right)$ . (6)

At time (t) is zero the deflection $\left(\theta \right)$ is also 0.

Substitute 0 for t and 0 for $\theta$.

$\begin{array}{c}0={\theta }_m\sin \left({\omega }_n\left(0\right)+\varphi \right)\\ \varphi =0\end{array}$

Differentiate the equation (6).

$\dot{\theta }={\theta }_m{\omega }_n\cos \left({\omega }_{n}t+\varphi \right)$

Substitute 0 for t and 0 for $\varphi$.

$\begin{array}{c}\dot{\theta }={\theta }_m{\omega }_n\cos \left({\omega }_n\left(0\right)+\left(0\right)\right)\\ ={\theta }_m{\omega }_n\end{array}$

Write the express the maximum velocity ${\left({\dot{x}}_A\right)}_m$ of vibration:

${\left({\dot{x}}_A\right)}_m=0.6l\left({\dot{\theta }}_m\right)$

Substitute ${\theta }_m{\omega }_n$ for $\dot{\theta }$.

${\left({\dot{x}}_A\right)}_m=0.6l\left({\theta }_m{\omega }_n\right)$ (7)

Calculate the angular amplitude $\left({\theta }_m\right)$:

Substitute $1.1\text{mm/s}$ for ${\left({\dot{x}}_A\right)}_m$, 600 mm for l, and $22.939\text{rad/s}$ for ${\omega }_n$ in equation (7).

$\begin{array}{l}1.1\text{mm/s}\times \frac{1\text{m}}{1000\text{mm}}=0.6\left(60\text{0}\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\left({\theta }_m\right)\left(22.939\text{rad/s}\right)\\ {\theta }_m=\frac{0.0011}{0.36\times 22.939}\\ {\theta }_m=1.332\times {10}^{-4}\text{rad}\times \frac{180°}{\pi }\\ {\theta }_m=0.0076°\end{array}$

Therefore, the amplitude $\left({\theta }_m\right)$ of the angular motion of the rod is $\underset{\_}{0.0076°}$.