Chapter 19.4, Problem 19.112P

To determine

The amplitude of vibration should be expected when three collar are placed on the spring.

Answer to Problem 19.112P

The amplitude of vibration of the system with three collars in out-phase motion is $-5.63\text{ mm}$ and the amplitude of vibration of the system with three collars in in-phase motion is $22.5\text{ mm}$.

Explanation of Solution

Given:

Amplitude of vibration when 1 collar is on the spring is $15\text{ mm}$.

Amplitude of vibration when 2 collar is on the spring is $18\text{ mm}$.

Concept used:

Write the expression for amplitude of forced vibration.

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

Here, $x_m$ is the amplitude of forced vibration, ${\delta }_{sr}$ is the static displacement, ${\omega }_f$ is the forced frequency and ${\omega }_n$ is the natural frequency of system.

Write the expression for natural circular frequency of system having mass $m$.

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

Here, ${\left({\omega }_n\right)}_1$ is the natural frequency of vibration with only one disk, $m$ is the mass of disk and $k$ is the stiffness of spring.

Write the expression for natural circular frequency of system having mass $2m$.

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

Substitute ${\left({\omega }_n\right)}_1$ for $\sqrt{\frac{k}{m}}$ in above expression.

${\left({\omega }_n\right)}_2=\frac{{\left({\omega }_n\right)}_1}{\sqrt{2}}$ ...... (3)

Here, ${\left({\omega }_n\right)}_2$ is the natural frequency of vibration with mass only two disks.

Write the expression for natural circular frequency of system having mass $3m$.

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

Substitute ${\left({\omega }_n\right)}_1$ for $\sqrt{\frac{k}{m}}$ in above expression.

${\left({\omega }_n\right)}_2=\frac{{\left({\omega }_n\right)}_1}{\sqrt{3}}$ ...... (4)

Here, ${\left({\omega }_n\right)}_3$ is the natural frequency of vibration with mass only three disks.

Substitute ${\left(x_m\right)}_1$ for $x_m$ and ${\left({\omega }_n\right)}_1$ in equation (1).

${\left(x_m\right)}_1=\frac{{\delta }_{st}}{1-\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (5)

Here, ${\left(x_m\right)}_1$ is the amplitude of vibration with 1 disk.

Substitute ${\left(x_m\right)}_2$ for $x_m$ and ${\left({\omega }_n\right)}_2$ in equation (1).

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

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

${\left(x_m\right)}_2=\frac{{\delta }_{st}}{1-2\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (6)

Here, ${\left(x_m\right)}_2$ is the amplitude of vibration with 2 disks

Substitute ${\left(x_m\right)}_3$ for $x_m$ and ${\left({\omega }_n\right)}_3$ in equation (1).

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

Substitute $\frac{{\left({\omega }_n\right)}_1}{\sqrt{3}}$ for ${\left({\omega }_n\right)}_2$ in above expression.

${\left(x_m\right)}_3=\frac{{\delta }_{st}}{1-3\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (7)

Here, ${\left(x_m\right)}_3$ is the amplitude of vibration with 3 disks.

Calculation:

In-Phase motion:

Substitute $15\text{ mm}$ for ${\left(x_m\right)}_1$ in equation (5).

$15\text{ mm}=\frac{{\delta }_{st}}{1-\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (8)

Substitute $18\text{mm}$ for ${\left(x_m\right)}_2$ in equation (6).

$18\text{ mm}=\frac{{\delta }_{st}}{1-2\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (9)

Divide equation (9) by equation (8).

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

Solve the above expression for $\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}$.

$\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}=\frac{1}{7}$

Substitute $\frac{1}{7}$ for $\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}$ in equation (8).

$\begin{array}{l}15=\frac{{\delta }_m}{1-\frac{1}{7}}\\ {\delta }_m=12.855\text{ mm}\end{array}$

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

$\begin{array}{c}{\left(x_m\right)}_3=\frac{12.855}{1-3\left(\frac{1}{7}\right)}\\ =22.5\text{ mm}\end{array}$

The amplitude of vibration of the system with three collars in in-phase motion is $22.5\text{ mm}$.

Out-of phase motion:

Substitute $-18\text{mm}$ for ${\left(x_m\right)}_2$ in equation (6).

$-18\text{ mm}=\frac{{\delta }_{st}}{1-2\left(\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}\right)}$ ...... (10)

Divide Equation (10) by equation (8).

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

Solve the above expression for $\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}$.

$\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}=0.647$

Substitute $0.647$ for $\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}$ in equation (8).

$\begin{array}{l}15=\frac{{\delta }_m}{1-0.647}\\ {\delta }_m=5.295\text{ mm}\end{array}$

Substitute $5.295\text{ mm}$ for ${\delta }_m$ and $0.647$ for $\frac{{\omega }_f^2}{{\left({\omega }_n\right)}_1^2}$ in equation (7).

$\begin{array}{c}{\left(x_m\right)}_3=\frac{5.295}{1-3\left(0.647\right)}\\ =-5.63\text{ mm}\end{array}$

The amplitude of vibration of the system with three collars in out-phase motion is $-5.63\text{ mm}$.

Conclusion:

Thus, the amplitude of vibration of the system with three collars in out-phase motion is $-5.63\text{ mm}$ and the amplitude of vibration of the system with three collars in in-phase motion is $22.5\text{ mm}$.