An industrial press is mounted on a rubber pad to isolate it from its foundation. If the rubber pad
is compressed 5 mm by the self weight of the press, find the natural frequency of the system.
$$\begin{align} & {{\delta }_{st}}=5\cdot {{10}^{-5}}m \\ & {{\omega }_{n}}=\sqrt{\frac{g}{{{\delta }_{st}}}}=44.2945\text{ rad/s=7}\text{.0497 Hz} \\ \end{align}$$
Example 2.2
A spring mass system
has a natural period of 0.5 s. Determine the value of new period if the spring
stiffness is decreased by 150 percent and increased by 1000 percent.
Solution: The formula
that gives connection between natural frequency and the period is:
$$\tau =\frac{2\pi }{{{\omega }_{n}}}=0.5$$
The expression for
natural frequency can be written in the following form:
$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$
Combining those two
expressions we get:
$$\begin{align}
& \tau =\frac{2\pi }{{{\omega }_{n}}}=\frac{2\pi }{\sqrt{\frac{k}{m}}}=2\pi \sqrt{\frac{m}{k}}=0.5s \\
& \sqrt{m}=0.5\frac{\sqrt{k}}{2\pi } \\
\end{align}$$
Example 2.3
A spring mass system
has the natural frequency of 2000 Hz. When we reduce the spring constant by
1000 N/mm the frequency is altered by 60 percent. Find the mass and the
stiffness of the system.
$$\begin{align}
& {{\omega }_{n}}=62.832\text{ rad/s=}\sqrt{\frac{k}{m}} \\
& \sqrt{m}=\frac{\sqrt{k}}{62.832} \\
& {{\omega }_{n1}}=0.55{{\omega }_{n}}=34.5576\text{ rad/s}=\sqrt{\frac{{{k}_{new}}}{{{m}_{new}}}}=\sqrt{\frac{k-800}{m}} \\
& 62.836\sqrt{\frac{k-800}{m}}=34.5576 \\
& \sqrt{\frac{k-800}{m}}=0.55 \\
& \frac{k-800}{m}={{0.55}^{2}}=0.3025 \\
& k=1146.9534\text{ N/m} \\
& \sqrt{m}=\frac{\sqrt{k}}{62.832} \\
& m=\frac{1146.9534}{3947.8602} \\
& m=0.2905\text{ kg} \\
\end{align}$$
Example 2.4
A helical spring which is fixed on the one end
and loaded on the other end, requires a force of 100 N to produce elongation of
10 mm. the ends of the spring are now rigidly fixed, one end vertically above
the other, and a block of mass 10 kg is attached at the middle point of its
length. Determine the time taken to complete one vibration cycle when the mass
is set vibrating in the vertical direction.
Example 2.5
An air-conditioning chiller unit weighing 9000 N is to be supported by four air springs. Design the air springs such that the natural frequency of vibration of the unit lies between 5 rad/s and 10 rad/s.
Example 2.6
The maximum velocity
attained by the mass of a simple harmonic oscillator is 10 cm/s, and
the period of oscillation is
2 s. If the mass is released with an initial displacement of 2 cm,
find (a) the amplitude, (b)
the initial velocity, (c) the maximum acceleration, and (d) the
phase angle.
$$\begin{align}
& x=A\cos ({{\omega }_{n}}t-{{\phi }_{0}}) \\
& \dot{x}=-A{{\omega }_{n}}\sin ({{\omega }_{n}}t-{{\phi }_{0}}) \\
& \ddot{x}=-A\omega _{n}^{2}\cos ({{\omega }_{n}}t-{{\phi }_{0}}) \\
& a){{\omega }_{n}}A=0.1\text{ m/s} \\
& {{\tau }_{n}}=\frac{2\pi }{{{\omega }_{n}}}=2s\Rightarrow {{\omega }_{n}}=3.1416\text{ rad/s} \\
& \text{A=}\frac{0.1}{{{\omega }_{n}}}=0.03183\text{ m} \\
& \text{d)}{{\text{x}}_{0}}=x(t=0)=Acos\left( -{{\phi }_{0}} \right)=0.02m \\
& cos\left( -{{\phi }_{0}} \right)=\frac{0.02}{A}=0.6283 \\
& {{\phi }_{0}}={{51.0724}^{\circ }} \\
& \text{b)}{{{\text{\dot{x}}}}_{0}}=\dot{x}\left( t=0 \right)=-{{\omega }_{n}}A\sin \left( -{{\phi }_{0}} \right)=-0.1\sin \left( -{{51.0724}^{\circ }} \right) \\
& {{{\dot{x}}}_{0}}=0.07779\text{ m/s} \\
& \text{c)}{{{\text{\ddot{x}}}}_{\max }}=\omega _{n}^{2}A={{3.1416}^{2}}\cdot 0.03183=00314151m/{{s}^{2}} \\
\end{align}$$
Example 2.7
Three springs and a mass are
attached to a rigid, weightless bar PQ as shown in next figure
Find the natural frequency
of vibration of the system.
Example 2.8
An automobile having a mass
of 2,000 kg deflects its suspension springs 0.02 m under static conditions.
Determine the natural frequency of the automobile in the vertical direction by assuming
damping to be negligible.
$$\begin{align}
& m=2000kg \\
& {{\delta }_{st}}=0.02m \\
& {{\omega }_{n}}=\sqrt{\frac{g}{{{\delta }_{st}}}}=\sqrt{\frac{9.81}{0.02}}=22.1472\text{ rad/s} \\
\end{align}$$
Example 2.9
Find the natural frequency
of vibration of a spring-mass system arranged on an inclined plane, as shown in next
figure.
$$\begin{align} & m\ddot{x}=-{{k}_{1}}\left( x+{{\delta }_{st}} \right)-{{k}_{2}}\left( x+{{\delta }_{st}} \right)+W\sin \theta \\ & {{\delta }_{st}}\left( {{k}_{1}}+{{k}_{2}} \right)=W\sin \theta \\ & m\ddot{x}+\left( {{k}_{1}}+{{k}_{2}} \right)x=0 \\ & {{\omega }_{n}}=\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}} \\ \end{align}$$
Example 2.10
A loaded mine cart weighing 20 KN is being
lifted by a frictionless pulley and a wire rope as shown in next figure. Find
the natural frequency of vibration of the cart in the given position.
$$\begin{align}
& {{k}_{1}}=\frac{{{A}_{1}}{{E}_{1}}}{{{l}_{1}}}=\frac{\frac{\pi }{4}\cdot {{(0.001)}^{2}}\left( 200\cdot {{10}^{9}} \right)}{9}=17453.3N/m \\
& {{k}_{2}}=\frac{{{A}_{2}}{{E}_{2}}}{{{l}_{2}}}=\frac{17453.3\left( 7.5 \right)}{9}=1454.4.3N/m \\
& {{k}_{eq}}={{k}_{1}}+{{k}_{2}}=31997.7N/m \\
& mx=-\left( {{k}_{1}}+{{k}_{2}} \right)\left( x+{{\delta }_{st}} \right)+W\sin \theta \\
& \left( {{k}_{1}}+{{k}_{2}} \right){{\delta }_{st}}=W\sin \theta \\
& m\ddot{x}+\left( {{k}_{1}}+{{k}_{2}} \right)x=0 \\
& \omega =\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}=3.9597\text{ rad/s} \\
\end{align}$$
Example 2.11
An electronic chassis weighing 500 N is
isolated by supporting it on four helical springs. Design the springs so that
the unit can be used in an environment in which the vibratory frequency ranges
from 0 to 5 Hz.
$$\begin{align}
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=62.832 \\
& {{k}_{eq}}=m\omega _{m}^{2}=\frac{500}{9.81}{{62.832}^{2}}=20.1857\cdot {{10}^{4}}N/m=4k \\
& k=50464.25N/m \\
& k=\frac{G{{d}^{4}}}{8n{{D}^{3}}} \\
& D=\sqrt[3]{\frac{8kn}{G{{d}^{4}}}}=0.0291492m=2.91492cm \\
\end{align}$$
Example 2.12
Find
the natural frequency of the system shown in next figure with and without the
springs k1 and
k2 in the middle of the elastic beam.
Example 2.13
Find
the natural frequency of the pulley system shown in next figure by neglecting
the friction
and
the masses of the pulleys.
Let x1,x2 represents
displacements of pulleys 1 and 2
x=x1+x2
Let P be the tension in
rope
For equilibrium of
pulley 1
2P=k*x1
For equilibrium of
pulley 2
2P=k*x2
$$\begin{align}
& \frac{1}{{{k}_{1}}}=\frac{1}{4k}+\frac{1}{4k}=\frac{1}{2k} \\
& {{k}_{1}}=2k \\
& {{k}_{2}}=k+k=2k \\
& x=2{{x}_{1}}+2{{x}_{2}}=2\left( \frac{2P}{{{k}_{1}}} \right)+2\left( \frac{2P}{{{k}_{2}}} \right) \\
& x=4P\left( \frac{1}{2k}+\frac{1}{2k} \right)=\frac{4P}{k} \\
& {{k}_{eq}}=\frac{P}{x}=\frac{k}{4} \\
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{k}{4m}} \\
\end{align}$$
Example 2.14
A
weight W is supported by three frictionless and massless pulleys and a
spring of stiffness k, as shown in next figure. Find the natural
frequency of vibration of weight W for small oscillations.
Example 2.15
A
rigid block of mass M is mounted on four elastic supports, as shown in
Fig. 2.58. A mass m drops from a height l and adheres to the
rigid block without rebounding. If the spring constant
of
each elastic support is k, find the natural frequency of vibration of
the system (a) without the mass m, and (b) with the mass m. Also
find the resulting motion of the system in case (b).
Example 2.16
A
sledgehammer strikes an anvil with a velocity of 20 m/sec. The hammer and the anvil
weigh 50 N and 500 N, respectively. The anvil is supported on four springs,
each of stiffness k=20 kn/m Find the resulting motion of the anvil (a) if the
hammer remains in contact with the anvil and (b) if the hammer does not remain in
contact with the anvil after the initial impact.
Solution:
a)Velocity
of hammer v= 20m/s, x=0 at static equilibrium postion
$$\begin{align}
& {{x}_{0}}=x(t=0)=-\frac{W}{{{k}_{eq}}}=-\frac{mg}{4k} \\
& \left( M+m \right){{{\dot{x}}}_{0}}=mv \\
& {{{\dot{x}}}_{0}}=\dot{x}(t=0)=\frac{mv}{M+m} \\
& {{\omega }_{n}}=\sqrt{\frac{4k}{M+m}} \\
& x\left( t \right)={{A}_{0}}\sin \left( {{\omega }_{n}}t+{{\phi }_{0}} \right) \\
& {{A}_{0}}={{\left[ x_{0}^{2}+{{\left( \frac{{{x}_{0}}}{{{\omega }_{n}}} \right)}^{2}} \right]}^{\frac{1}{2}}}={{\left[ \frac{{{m}^{2}}{{g}^{2}}}{16{{k}^{2}}}+\frac{{{m}^{2}}{{v}^{2}}}{\left( M+m \right)4k} \right]}^{\frac{1}{2}}}=0.04816\text{ m} \\
& {{\phi }_{0}}={{\tan }^{-1}}\left( \frac{g\sqrt{M+m}}{v\sqrt{4k}} \right)=-0.7436\deg \\
\end{align}$$
b)x=0
at static equilibrium position:
$$\begin{align}
& {{x}_{0}}=0 \\
& M{{{\dot{x}}}_{0}}=mv \\
& x(t)={{A}_{0}}\sin \left( {{\omega }_{n}}t+{{\phi }_{0}} \right) \\
& {{A}_{0}}={{\left[ x_{0}^{2}+{{\left( \frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}} \right)}^{2}} \right]}^{\frac{1}{2}}}=0.0505\text{ m} \\
& \phi {}_{0}={{\tan }^{-1}}\left[ \frac{{{x}_{0}}{{\omega }_{n}}}{{{{\dot{x}}}_{0}}} \right]={{\tan }^{-1}}(0)=0 \\
\end{align}$$
Example 2.17
Derive
the expression for the natural frequency of the system shown in next figure.
Note that
the
load W is applied at the tip of beam 1 and midpoint of beam 2.
Example 2.18
A
heavy machine weighing 9,810 N is being lowered vertically down by a winch at a
uniform velocity of 2 m/s. The steel cable supporting the machine has a
diameter of 0.01 m. The winch is suddenly stopped when the steel cable s length
is 20 m. Find the period and amplitude of the ensuing vibration of the machine.
$$\begin{align}
& k=\frac{AE}{l}=\frac{\left\{ \frac{\pi }{4}\cdot {{0.01}^{2}} \right\}\left\{ 2.07\cdot {{10}^{11}} \right\}}{20}=0.8129\cdot {{10}^{6}}\text{ N/m} \\
& m=1000\text{ kg} \\
& {{\omega }_{n}}=\sqrt{\frac{k}{m}}={{\left( \frac{0.8129\cdot {{10}^{6}}}{1000} \right)}^{\frac{1}{2}}}=28.5114\text{ rad/s} \\
& {{{\dot{x}}}_{0}}=2m/s \\
& {{x}_{0}}=0 \\
& {{\tau }_{n}}=\frac{2\pi }{{{\omega }_{n}}}=\frac{2\pi }{28.5114}0.2204\text{ s} \\
& A=\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}=\frac{2}{28.5114}=0.07015\text{ m} \\
\end{align}$$
Example 2.19
The natural frequency
of a spring-mass system is found to be 2 Hz. When an additional mass of 1 kg is
added to the original mass m, the natural frequency is reduced to 1 Hz. Find
the spring constant k and the mass m.
$$\begin{align}
& {{\omega }_{n}}=2Hz=12.5664\text{ rad/s}=\sqrt{\frac{k}{m}} \\
& \sqrt{k}=12.5664\sqrt{m} \\
& {{\omega }_{n}}'=\sqrt{\frac{k'}{m'}}=\sqrt{\frac{k}{m+1}}=6.2832\text{ rad/s} \\
& \sqrt{k}=6.2832\sqrt{m+1} \\
& \sqrt{k}=12.5664\sqrt{m} \\
& \sqrt{m+1}=2\sqrt{m} \\
& m=\frac{1}{3}kg \\
& k={{\left( 12.5664 \right)}^{2}}m=52.6381\text{ N/m} \\
\end{align}$$
Example 2.20
An electrical switch
gear is supported by a crane through a steel cable of length 4 m and diameter
0.01 m. If the natural time period of axial vibration of the switch gear is found
to be 0.1 s, find the mass of the switch gear.
$$\begin{align}
& k=\frac{AE}{l}=\frac{1}{4}\left( \frac{\pi }{4}{{0.01}^{2}} \right)2.07\cdot {{10}^{11}}=4.0644\cdot \text{1}{{\text{0}}^{6}}\text{N/m} \\
& {{\tau }_{n}}=0.1=\frac{1}{{{f}_{n}}}=\frac{2\pi }{{{\omega }_{n}}} \\
& {{\omega }_{n}}=\frac{2\pi }{0.1}=20\pi =\sqrt{\frac{k}{m}} \\
& m=\frac{k}{\omega _{n}^{2}}=\frac{4.0644\cdot \text{1}{{\text{0}}^{6}}}{{{\left( 20\pi \right)}^{2}}}=1029.53\text{ kg} \\
\end{align}$$
Example 2.21
Four
weightless rigid links and a spring are arranged to support a weight W in two
different ways, as shown in. Determine the natural frequencies of vibration of
the two arrangements.
$$\begin{align}
& b=2l\sin \theta \\
& a){{k}_{eq}}=k\left( \frac{4{{l}^{2}}-{{b}^{2}}}{{{b}^{2}}} \right)=k\left( \frac{4{{l}^{2}}-4{{l}^{2}}{{\sin }^{2}}\theta }{4{{l}^{2}}{{\sin }^{2}}\theta } \right) \\
& {{k}_{eq}}=k\left( \frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \right) \\
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{kg{{\cot }^{2}}\theta }{W}} \\
& b){{\omega }_{n}}=\sqrt{\frac{kg}{W}}\Rightarrow {{k}_{eq}}=k \\
\end{align}$$
Example 2.22
A
scissors jack is used to lift a load W. The links of the jack are rigid and the
collars can slide freely on the shaft against the springs of stiffness’s k1 and
k2. Find the natural frequency of vibration of the weight in the vertical
direction.
$$\begin{align}
& y=\sqrt{{{l}^{2}}-\left( l\sin \theta -x \right){}^{2}}-l\cos \theta \\
& y=\sqrt{{{l}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-\left( l\sin \theta -x \right){}^{2}}-l\cos \theta \\
& y=l\cos \theta \sqrt{1-\frac{{{x}^{2}}}{{{l}^{2}}{{\cos }^{2}}\theta }+\frac{2lx\sin \theta }{{{l}^{2}}{{\cos }^{2}}\theta }}-l\cos \theta \\
& \frac{1}{2}{{k}_{eq}}{{x}^{2}}=\frac{1}{2}{{k}_{1}}{{y}^{2}}+\frac{1}{2}{{k}_{2}}{{y}^{2}} \\
& y=l\cos \theta \left( 1-\frac{1}{2}\frac{{{x}^{2}}}{{{l}^{2}}{{\cos }^{2}}\theta }+\frac{1}{2}\frac{2lx\sin \theta }{{{l}^{2}}{{\cos }^{2}}\theta } \right)-l\cos \theta \\
& y=\frac{x\sin \theta }{\cos \theta }=x\tan \theta \\
& {{k}_{eq}}=\left( {{k}_{1}}+{{k}_{2}} \right){{\tan }^{2}}\theta \\
& m\ddot{x}+{{k}_{eq}}x=0 \\
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{\left( {{k}_{1}}+{{k}_{2}} \right)}{W}}\tan \theta \\
\end{align}$$
Example 2.23
A
weight is suspended using six rigid links and two springs in two different
ways, as shown in next figure. Find the natural frequencies of vibration of the two arrangements.
$$\begin{align}
& a){{k}_{eq}}=\frac{k}{2} \\
& m\ddot{x}+{{k}_{eq}}x=0 \\
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{k}{2m}} \\
& b){{x}_{s}}=x\frac{2}{b}\sqrt{{{l}^{2}}-\frac{{{b}^{2}}}{4}} \\
& \frac{1}{2}{{k}_{eq}}{{x}^{2}}=2\left( \frac{1}{2}kx_{s}^{2} \right) \\
& {{k}_{eq}}=2k{{\left( \frac{{{x}_{s}}}{k} \right)}^{2}}=2k\left( \frac{4}{{{b}^{2}}} \right)\left( {{l}^{2}}-\frac{{{b}^{2}}}{4} \right) \\
& {{k}_{eq}}=\frac{8k}{{{b}^{2}}}\left( {{l}^{2}}-\frac{{{b}^{2}}}{4} \right) \\
& m\ddot{x}+{{k}_{eq}}x=0 \\
& {{\omega }_{n}}=\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{8k}{{{b}^{2}}m}\left( {{l}^{2}}-\frac{{{b}^{2}}}{4} \right)} \\
\end{align}$$
Example 2.24
Next figure shows a
small mass m restrained by four linearly elastic springs, each of which has an
un-stretched length l, and an angle of orientation of 45 with respect to the
x-axis. Determine the equation of motion for small displacements of the mass in
the x direction.
$$\begin{align}
& {{F}_{1}}={{F}_{3}}={{k}_{1}}x\cos 45 \\
& {{F}_{2}}={{F}_{4}}={{k}_{2}}x\sin 135 \\
& F={{F}_{1}}\cos 45+{{F}_{2}}\cos 135+{{F}_{3}}\cos 45+{{F}_{4}}\cos 135 \\
& F=2x({{k}_{1}}co{{s}^{2}}45+{{k}_{2}}co{{s}^{2}}135) \\
& {{k}_{eq}}=\frac{F}{x}=2\left( \frac{{{k}_{1}}}{2}+\frac{{{k}_{2}}}{2} \right)={{k}_{1}}+{{k}_{2}} \\
& m\ddot{x}+({{k}_{1}}+{{k}_{2}})x=0 \\
\end{align}$$
thinks
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