Example 1
A weight of 50 N is
suspended from a spring of stiffness 4000 N/m and is subjected to a harmonic force
of amplitude 60 N and frequency 6 Hz. Find (a) the extension of the spring due to
the suspended weight, (b) the static displacement of the spring due to the
maximum applied force, and (c) the amplitude of forced motion of the weight.
$$\begin{align}
& a)\delta =\frac{W}{k}=\frac{50}{4000}=0.0125\text{m} \\
& b){{\delta }_{st}}=\frac{{{F}_{0}}}{k}=\frac{60}{4000}=0.015\text{m} \\
& c){{\omega }_{n}}=\sqrt{\frac{k}{m}}=\sqrt{\frac{4000\cdot 9.81}{50}}=28.0143\text{rad/s} \\
& \omega =6\text{Hz}=37.6992\text{rad/s} \\
& \text{X=}{{\delta }_{st}}\left| \frac{1}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right|=0.015\left| \frac{1}{1-{{\left( \frac{37.6992}{28.0143} \right)}^{2}}} \right|=0.0185\text{m} \\
\end{align}$$
Example 2
A spring-mass system is
subjected to a harmonic force whose frequency is close to the natural frequency
of the system. If the forcing frequency is 39.8 Hz and the natural frequency is
40.0 Hz, determine the period of beating
$${{\tau }_{b}}=\frac{2\pi }{{{\omega }_{n}}-\omega }=\frac{2\pi }{2\pi \left( 40-39.8 \right)}=5\text{ s}$$
Example 3
Consider a spring-mass
system, with k=4000N/m and m=10kg subject
to a harmonic force F(t)=400cos10t N.Find and plot the total response of the
system under the following initial conditions: a)x0=0.1 m, x0’=0,b)x0=0,x0’=10m/s,c)x0=0.1
m,x0’=10m/s
$$\begin{align}
& k=4000\text{ N/m} \\
& m=10\text{ kg} \\
& F(t)=400\cos 10t\text{ N} \\
& {{F}_{0}}=400\text{ N} \\
& \omega =10\text{ rad/s} \\
& {{\omega }_{n}}=\sqrt{\frac{k}{m}}=20\text{ rad/s} \\
& \frac{\omega }{{{\omega }_{n}}}=\frac{10}{20}=0.5$$
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Example 3
Consider a spring-mass
system, with k=4000N/m and m=10kg subject
to a harmonic force F(t)=400cos20t N. Find and plot the total response of the
system under the following initial conditions: a)x0=0.1 m, x0’=0,b)x0=0,x0’=10m/s,
c)x0=0.1 m,x0’=10m/s
$$\begin{align}
& k=4000\text{ N/m} \\
& m=10\text{ kg} \\
& F(t)=400\cos 20t\text{ N} \\
& {{F}_{0}}=400\text{ N} \\
& \omega =20\text{ rad/s} \\
& {{\omega }_{n}}=\sqrt{\frac{k}{m}}=20\text{ rad/s} \\
& \frac{\omega }{{{\omega }_{n}}}=\frac{20}{20}=1 \\
& x\left( t \right)={{x}_{0}}\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+\frac{{{\delta }_{st}}{{\omega }_{n}}t}{2}\sin {{\omega }_{n}}t \\
& {{\delta }_{st}}=\frac{{{F}_{0}}}{k}=\frac{400}{4000}=0.1 \\
& a){{x}_{0}}=0.1,{{{\dot{x}}}_{0}}=0 \\
& x(t)=0.1\cos 20t+\frac{0.1\cdot 20t}{2}\sin 20t \\
& x(t)=0.1\cos 20t+t\sin 20t \\
& b){{x}_{0}}=0,{{{\dot{x}}}_{0}}=10 \\
& x(t)=\frac{10}{20}\sin 20t+\frac{0.1\cdot 20t}{2}\sin 20t \\
& x(t)=0.5\sin 20t+t\sin 20t \\
& c){{x}_{0}}=0.1,{{{\dot{x}}}_{0}}=10 \\
& x(t)=0.1\cos 20t+\frac{10}{20}\sin 20t+\frac{0.1\cdot 20t}{2}\sin 20t \\
& x(t)=0.1\cos 20t+0.5\sin 20t+t\sin 20t \\
\end{align}$$
Example 3
Consider a spring-mass
system, with k=4000N/m and m=10kg subject
to a harmonic force F(t)=400cos20.1t N. Find and plot the total response of the
system under the following initial conditions: a)x0=0.1 m, x0’=0,b)x0=0,x0’=10m/s,
c)x0=0.1 m,x0’=10m/s
$$\begin{align}
& k=4000N/m \\
& m=10kg \\
& F(t)=400\cos 20.1tN \\
& {{F}_{0}}=400N \\
& \omega =20.1rad/s \\
& \underline{{{\omega }^{2}}=404.01} \\
& {{\omega }_{n}}=\sqrt{\frac{k}{m}}=20rad/s \\
& x\left( t \right)=\left( {{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+\left( \frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t \\
& a){{x}_{0}}=0.1,{{{\dot{x}}}_{0}}=0 \\
& x(t)=\left\{ 0.1-\frac{400}{4000-10(404.01)} \right\}\cos 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\
& x(t)=10.075062\cos 20t-9.975062\cos 20.1t \\
& b){{x}_{0}}=0,{{{\dot{x}}}_{0}}=10 \\
& x(t)=-\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20t+\frac{10}{20}\sin 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\
& x(t)=9.975062\cos 20t+0.5\sin 20t-9.975062\cos 20.1t \\
& c){{x}_{0}}=0,{{{\dot{x}}}_{0}}=10 \\
& x(t)=\left\{ 0.1-\frac{400}{4000-10(404.01)} \right\}\cos 20t+\frac{10}{20}\sin 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\
& x(t)=10.075062\cos 20t+0.5\sin 20t-9.975062\cos 20.1t \\
\end{align}$$
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