Harmonically excited vibration example 1-5

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

$$\begin{align} & k=4000\text{ N/m} \\ & m=10\text{ kg} \\ & F(t)=400\cos 10t\text{ N} \\ & {{F}_{0}}=400\text{ N} \\ & \omega =10\text{ rad/s} \\ & {{\omega }_{n}}=\sqrt{\frac{k}{m}}=20\text{ rad/s} \\ & \frac{\omega }{{{\omega }_{n}}}=\frac{10}{20}=0.5$$ <1 0.1-="" 10t="" 20.1t="" 20t="" a="" align="" b="" c="" cos10t="" cos="" dot="" end="" frac="" k-m="" left="" n="" omega="" right="" sin="" span="" style="font-family: "times new roman", serif; line-height: 115%;" t="" x="">

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

$$\begin{align} & k=4000N/m \\ & m=10kg \\ & F(t)=400\cos 20.1tN \\ & {{F}_{0}}=400N \\ & \omega =20.1rad/s \\ & \underline{{{\omega }^{2}}=404.01} \\ & {{\omega }_{n}}=\sqrt{\frac{k}{m}}=20rad/s \\ & x\left( t \right)=\left( {{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+\left( \frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t \\ & a){{x}_{0}}=0.1,{{{\dot{x}}}_{0}}=0 \\ & x(t)=\left\{ 0.1-\frac{400}{4000-10(404.01)} \right\}\cos 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\ & x(t)=10.075062\cos 20t-9.975062\cos 20.1t \\ & b){{x}_{0}}=0,{{{\dot{x}}}_{0}}=10 \\ & x(t)=-\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20t+\frac{10}{20}\sin 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\ & x(t)=9.975062\cos 20t+0.5\sin 20t-9.975062\cos 20.1t \\ & c){{x}_{0}}=0,{{{\dot{x}}}_{0}}=10 \\ & x(t)=\left\{ 0.1-\frac{400}{4000-10(404.01)} \right\}\cos 20t+\frac{10}{20}\sin 20t+\left\{ \frac{400}{4000-10(404.01)} \right\}\cos 20.1t \\ & x(t)=10.075062\cos 20t+0.5\sin 20t-9.975062\cos 20.1t \\ \end{align}$$