Chapter 19.4, Problem 19.112P

Rod AB is rigidly attached to the frame of a motor running at a constant speed. When a collar of mass m is placed on the spring, it is observed to vibrate with an amplitude of 15 mm. When two collars, each of mass m, are placed on the spring, the amplitude is observed to be 18 mm. What amplitude of vibration should be expected when three collars, each of mass m, are placed on the spring? (Obtain two answers.)

Fig. P19.112

To determine

Find the amplitude of vibration ${\left(x_m\right)}_3$ should be expected when three collars are placed on the spring.

Answer to Problem 19.112P

The amplitude of vibration ${\left(x_m\right)}_3$ is $\underset{\_}{22.5\text{mm}\text{for}x>0\text{and }-\text{5}\text{.63}\text{mm}\text{for}x<0}$.

Explanation of Solution

Given information:

The amplitude of the one collar ${\left(x_m\right)}_1$ is 15 mm.

The amplitude of the two collars ${\left(x_m\right)}_2$ is 18 mm.

Calculation:

The expression for the natural frequency $\left({\omega }_n\right)$ as follows:

${\omega }_n^2=\frac{k}{m}$ (1)

Here, ${\omega }_n$ is the natural circular frequency, k is the stiffness or spring constant and m is the mass of the collar.

The expression for the amplitude of forced vibration $\left(x_m\right)$ as follows:

$x_m=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\omega }_n}\right)}^2}$ (2)

Here, $x_m$ is the amplitude of forced vibration, ${\delta }_m$ is the maximum amplitude, and ${\omega }_f$ is the frequency of periodic force.

Consider only one collar is placed.

Calculate the natural frequency when only one collar is placed using the relation:

${\left({\omega }_n^2\right)}_1=\frac{k}{m}$ (3)

Here, ${\left({\omega }_n^2\right)}_1$ the square of the natural frequency when only one collar is placed.

Substitute ${\left({\omega }_n\right)}_1$ for ${\omega }_n$ in Equation (2).

${\left(x_m\right)}_1=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}$

Here, ${\left(x_m\right)}_1$ is the amplitude of the system with one collar.

Substitute 15 mm for ${\left(x_m\right)}_1$.

$15\text{mm}=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}$ (4)

Consider two collars are placed.

Find the natural frequency:

Substitute 2m for m in equation (1).

${\left({\omega }_n^2\right)}_2=\frac{k}{2m}$ (5)

Here, ${\left({\omega }_n^2\right)}_2$ is the square of the natural frequency when two collars are placed.

Substitute $\frac{k}{m}$ for ${\left({\omega }_n^2\right)}_1$.

$\begin{array}{l}{\left({\omega }_n^2\right)}_2=\frac{1}{2}\frac{k}{m}\\ {\left({\omega }_n^2\right)}_2=\frac{1}{2}{\left({\omega }_n^2\right)}_1\\ {\left({\omega }_n\right)}_2=\frac{1}{\sqrt{2}}{\left({\omega }_n\right)}_1\end{array}$

Multiply both sides by ${\omega }_f$.

$\begin{array}{l}{\omega }_f{\left({\omega }_n\right)}_2={\omega }_f\frac{1}{\sqrt{2}}{\left({\omega }_n\right)}_1\\ \sqrt{2}\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}=\frac{{\omega }_f}{{\left({\omega }_n\right)}_2}\\ \frac{{\omega }_f}{{\left({\omega }_n\right)}_2}=\sqrt{2}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)\end{array}$ (6)

Substitute ${\left({\omega }_n\right)}_2$ for ${\omega }_n$ in equation (2).

${\left(x_m\right)}_2=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_2}\right)}^2}$ (7)

Here, ${\left(x_m\right)}_2$ is the amplitude of the system with two collar.

Substitute $\sqrt{2}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)$ for $\frac{{\omega }_f}{{\left({\omega }_n\right)}_2}$.

$\begin{array}{c}{\left(x_m\right)}_2=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_2}\right)}^2}\\ =\frac{{\delta }_m}{1-{\left(\sqrt{2}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)\right)}^2}\\ =\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\end{array}$ (8)

Consider three collars are placed.

Find the natural frequency:

Substitute 3m for m in equation (1).

${\left({\omega }_n^2\right)}_3=\frac{k}{3m}$ (9)

Here, ${\left({\omega }_n^2\right)}_3$ is the square of the natural frequency when three collars are placed.

Substitute $\frac{k}{m}$ for ${\left({\omega }_n^2\right)}_1$.

$\begin{array}{l}{\left({\omega }_n^2\right)}_3=\frac{1}{3}\frac{k}{m}\\ {\left({\omega }_n^2\right)}_3=\frac{1}{3}{\left({\omega }_n^2\right)}_1\\ {\left({\omega }_n\right)}_3=\frac{1}{\sqrt{3}}{\left({\omega }_n\right)}_1\end{array}$

Multiply both sides by ${\omega }_f$.

$\begin{array}{l}{\omega }_f{\left({\omega }_n\right)}_3={\omega }_f\frac{1}{\sqrt{3}}{\left({\omega }_n\right)}_1\\ \sqrt{3}\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}=\frac{{\omega }_f}{{\left({\omega }_n\right)}_3}\\ \frac{{\omega }_f}{{\left({\omega }_n\right)}_3}=\sqrt{3}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)\end{array}$ (10)

Substitute ${\left({\omega }_n\right)}_3$ for ${\omega }_n$ in equation (2).

${\left(x_m\right)}_3=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_3}\right)}^2}$ (11)

Here, ${\left(x_m\right)}_3$ is the amplitude of the system with three collars.

Substitute $\sqrt{3}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)$ for $\frac{{\omega }_f}{{\left({\omega }_n\right)}_3}$.

$\begin{array}{c}{\left(x_m\right)}_3=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_3}\right)}^2}\\ =\frac{{\delta }_m}{1-{\left(\sqrt{3}\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)\right)}^2}\\ =\frac{{\delta }_m}{1-3{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\end{array}$ (12)

The amplitude given in equation (8), can be in-phase with or out-of-phase with the periodic force.

In-phase motion:

Substitute 18 mm for ${\left(x_m\right)}_2$ in equation (8).

$\begin{array}{l}{\left(x_m\right)}_2=\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ 18\text{mm}=\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\end{array}$ (13)

Divide equation (13) by equation (4).

$\begin{array}{l}\frac{18\text{mm}}{15\text{mm}}=\frac{\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}}{\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}}\\ 1.2=\frac{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ 1.2\left(1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\right)=1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\\ 1.2-2.4{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\end{array}$

$\begin{array}{l}1.2-1=\left(2.4-1\right){\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\\ {\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=\frac{0.2}{1.4}\\ {\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=\frac{1}{7}\end{array}$

Substitute $1/7$ for ${\left({\omega }_f/{\left({\omega }_n\right)}_1\right)}^2$ in equation (4).

$\begin{array}{l}15\text{mm}=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ 15=\frac{{\delta }_m}{1-\frac{1}{7}}\\ 15=\frac{{\delta }_m}{0.857}\\ {\delta }_m=12.855\text{mm}\end{array}$

Substitute 12.855 mm for ${\delta }_m$ and $1/7$ for ${\left({\omega }_f/{\left({\omega }_n\right)}_1\right)}^2$ in equation (12).

$\begin{array}{c}{\left(x_m\right)}_3=\frac{{\delta }_m}{1-3{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ =\frac{12.855\text{mm}}{1-3\left(\frac{1}{7}\right)}\\ =\frac{12.855}{0.57143}\\ =22.5\text{mm}\end{array}$

Out-of-phase motion:

Substitute -18 mm for ${\left(x_m\right)}_2$ in equation (8).

$\begin{array}{l}{\left(x_m\right)}_2=\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ -18\text{mm}=\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\end{array}$ (14)

Divide equation (14) by equation (4).

$\begin{array}{l}\frac{-18\text{mm}}{15\text{mm}}=\frac{\frac{{\delta }_m}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}}{\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}}\\ -1.2=\frac{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}{1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ -1.2\left(1-2{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\right)=1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\\ -1.2+2.4{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2\end{array}$

$\begin{array}{l}\left(2.4+1\right){\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=1.2+1\\ {\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=\frac{2.2}{3.4}\\ {\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2=0.647\end{array}$

Substitute 0.647 for ${\left({\omega }_f/{\left({\omega }_n\right)}_1\right)}^2$ in equation (4).

$\begin{array}{l}15\text{mm}=\frac{{\delta }_m}{1-{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ 15=\frac{{\delta }_m}{1-0.647}\\ 15=\frac{{\delta }_m}{0.353}\\ {\delta }_m=5.295\text{mm}\end{array}$

Substitute 5.295 mm for ${\delta }_m$ and 0.64706 for ${\left({\omega }_f/{\left({\omega }_n\right)}_1\right)}^2$ in equation (12).

$\begin{array}{c}{\left(x_m\right)}_3=\frac{{\delta }_m}{1-3{\left(\frac{{\omega }_f}{{\left({\omega }_n\right)}_1}\right)}^2}\\ =\frac{5.295\text{mm}}{1-3\left(0.647\right)}\\ =\frac{5.295}{-0.941}\\ =-5.63\text{mm}\end{array}$

Therefore, the amplitude of vibration ${\left(x_m\right)}_3$ is $\underset{\_}{22.5\text{mm}\text{for}x>0\text{and }-\text{5}\text{.63}\text{mm}\text{for}x<0}$.