Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Answer to Problem 13.1P

The steady state amplitude of motion of the mass is $X=6.20\text{mm}$.

Explanation of Solution

Given:

$\begin{array}{l}m\text{=0}\text{.5}\text{kg}\\ k\text{=500}\text{N/m}\\ Y\text{=4}\text{mm}\\ f\text{=3}\text{Hz}\end{array}$

Concept Used:

Displacement transmissibility.

Calculation:

The equation of motion of mass is given as:

$m\stackrel{••}{x}+kx=ky$

Where,

$\begin{array}{l}y(t)=Y\sin \omega t\\ \omega =2\pi f\\ f\text{=3}\text{Hz}\\ \omega =6\pi \text{rad/sec}\end{array}$

Substitute the value in the equation,

We get,

$y(t)=\text{4}\times {\text{10}}^{-3}\sin 6\pi t$

Natural frequency is given as

$\begin{array}{c}{\omega }_n=\sqrt{\frac{k}{m}}\\ {\omega }_n{}^2=\frac{k}{m}\\ {\omega }_n{}^2=\frac{\text{500}}{\text{0}\text{.5}}\\ {\omega }_n=\sqrt{1000}\text{rad/sec}\\ \text{=31}\text{.62}\text{rad/sec}\end{array}$

Frequency ratio is given as:

$r\text{=}\frac{\omega }{{\omega }_n}$

Squaring both sides, we get,

$\begin{array}{l}{r}^2\text{=}\frac{{\left(6\pi \right)}^2}{1000}\\ r^2\text{=}0.3553\end{array}$

Displacement transmissibility for base excitation is given as:

$\frac{X}{Y}=\sqrt{\frac{4{\zeta }^2{r}^2+1}{{(1-r^2)}^2+4{\zeta }^2{r}^2}}$

Steady state amplitude of motion of $X$ is given as:

$X=Y\sqrt{\frac{4{\zeta }^2{r}^2+1}{{(1-r^2)}^2+4{\zeta }^2{r}^2}}$

Neglect the damping, $\zeta =0$

Replace the value of $\zeta =0$ in the above equation, we get

$\begin{array}{l}X=Y\sqrt{\frac{0+1}{{(1-r^2)}^2+0}}\\ X=Y\sqrt{\frac{1}{{(1-r^2)}^2}}\\ X=Y\frac{1}{(1-r^2)}\end{array}$

Substitute $r^2\text{=}0.3553$ and $Y=\text{4}\times {\text{10}}^{-3}$

$\begin{array}{l}X=\text{4}\times {\text{10}}^{-3}\frac{1}{(1-0.3553)}\\ X=6.20\times {10}^{-3}\text{m}\\ X=6.20\text{mm}\end{array}$

Conclusion:

The steady state amplitude of motion of the mass is $X=6.20\text{mm}$.