Free and Forced Vibrations
Free Vibrations
Learning Objectives:

Calculate the natural frequency vibration for an undamped free vibration.

Understand what do you mean by damping and its effect in a springdamper mass system vibrations.

Identify whether an damped free vibration is underdamped, critically damped or overdamped.
Types of Vibrations:
There are three important types of vibrations from subject point of view:

Free or natural vibrations : Natural vibrations take place at a frequency called fundamental or natural frequency, which is characteristics of the given component based on its material and geometrical properties. Free vibrations lack damping, dissipating or viscous forces. Applied force is applied only at the beginning of the motion and it is not continued further.

Damped vibrations : Here, the amplitude of the vibrations keeps on reducing every cycle of vibration. This is because energy is lost to the surroundings in terms of friction, viscous resistance or hysteresis. These losses are irreversible, the system energy reduces every oscillation and hence the amplitude of vibrations goes on reducing.

Forced Vibrations : Here the body is periodically subjected to disturbing forces, and it continues vibrating with the frequency other than the natural frequency of the component.
Undamped Free Vibration
\[m \ddot{x} = mg  k( x + \delta_{st})\]
since \[mg = k \delta_{st}\] \[m \ddot{x} =  kx\] or \[m \ddot{x} + kx = 0\] \[\ddot{x} + \frac{k}{m} x = 0\] The solution of this differential equation is:
where \[\omega_n = \sqrt{\frac{k}{m}}= \mathrm{natural\ frequency}\]
\(C_1\) and \(C_2\) are determined from the boundary (initial) conditions.
Damped Free Vibrations:
A damper generally consists of a plunger inside an oil filled cylinder, which dissipates energy by churning the oil.
When damping is present, the equation of motion is: \[m\ddot{x} + c\dot{x} + kx = 0\]
Case I. Two distinct (negative) real roots
If \(c^2 > 4mk\), the system is overdamped.
\[y = C_1 e^{r_1t} + C_2 e^{r_2t}\] Case II. One repeated (negative) real root
If \(c^2 = 4mk\), the system is critically damped.
\[y = C_1 e^{rt} + C_2 t e^{rt}\] Case III. Two complex conjugate roots
If \(c^2 < 4mk\), the system is underdamped. \[y = C_1 e^{At} \cos Bt + C_2 e^{At} \sin Bt\] where A is the real part of the root of characteristics equation, and B is its imaginary part.
Solved Example: 38101
In which type of vibrations, amplitude of vibration goes on decreasing every cycle? (Based on TANGEDCO AE ME 2015)
A. Damped vibrations
B. Undamped vibrations
C. Both (A) and (B)
D. None of the above
Damping of vibrations results in dissipation of energy due to friction, noise or unwanted signals, thereby reduction in amplitude of oscillatory motion.
Correct Answer: A
Solved Example: 38102
The natural frequency of a spring mass system on earth is $\omega_n$ The natural frequency of this system on the moon ($g_{moon}$ = $\dfrac{g_{earth}}{6}$ )will be: (GATE ME 2010)
A. $\omega_n$
B. 0.408 $\omega_n$
C. 0.204 $\omega_n$
D. 0.167 $\omega_n$
The natural frequency $\omega_n = \sqrt{\dfrac{k}{m}}$ does not depend upon gravity.
Correct Answer: A
Solved Example: 38103
The damping ratio of single DOF spring mass damping system, with mass of 1 kg, stiffness = 100 N/m and viscous damping coefficient of 25 Ns/m is: (GATE ME 2014 Shift 3)
A. 1.3
B. 1.4
C. 1.25
D. 1.2
Damping ratio, \begin{align*} \xi &= \dfrac{C}{C_{c}} = \dfrac{25}{2\sqrt{1 * 100}} =\dfrac{25}{20} =1.25 \end{align*}
Correct Answer: C
Solved Example: 38104
A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10$^{4}$ seconds. The amplitude in mm of the resulting free vibration is: (GATE ME 2013)
A. 10.0
B. 0.5
C. 1.0
D. 5.0
Given mass is initially at rest, u=0
We know that $F = ma$
\[a = \dfrac{F}{m} =\dfrac{5000}{1} =5000\ m/s^2\] \[v = u + at =0 + (5000*0.0001) = 0.5 m/s\] \begin{align*} KE_{max} = PE_{max}\\ \dfrac{1}{2}(m v^2) = \dfrac{1}{2}(k x^2)\\ x=5\ mm\\ \end{align*}Correct Answer: D
Solved Example: 38105
A mass of 1 kg is suspended by means of three springs as shown in the figure. The natural frequency of the system is approximately:
A. 46.90 Hz
B. 52.44 Hz
C. 77.46 Hz
D. 60.55 Hz
The series equivalent of spring combination will yield: \[ \dfrac{1}{k} = \dfrac{1}{1} + \dfrac{1}{3}, \ \mathrm{Or,\ }k = \dfrac{3}{4}\] The parallel combination of springs will yield: \[ k = k_1 + k_2\] \[ k = \dfrac{3}{4} + 2 = \dfrac{11}{4} kN/m = 2750 N/m\] The natural frequency will be given by: $\omega = \sqrt{\dfrac{k}{m}} = 52.54 Hz$
Correct Answer: B
Solved Example: 38106
Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in $N.m^2$) of the beam is: (GATE ME 2014)
A. 0.01
B. 0.065
C. 0.09
D. 0.12
\[S=\dfrac {FL^{3}}{3EI}\] \[k=\dfrac {F}{s}=\dfrac {3EI}{l^{3}} =\dfrac {3EI}{0.01^{3}} =3\times 10^{6}EI\] \[w_{n} =\sqrt {\dfrac {k}{m}} =\sqrt {\dfrac {3\times 10^{6}EI}{5}}w_{n} =2449.48\sqrt {EI}\] \begin{align*} f_{n} &= \dfrac {\omega _{n}}{2\pi}\\ 100 &= \dfrac {2449.48\sqrt {EI}}{2\pi}\\ EI &= 0.065N.m^{2} \end{align*}
Correct Answer: B
Solved Example: 38107
Considering massless rigid rod and small oscillations, the natural frequency (in rad/s) of vibration of the system shown in the figure is: (GATE ME 2015)
A. $\sqrt{\dfrac{400}{1}}$
B. $\sqrt{\dfrac{400}{2}}$
C. $\sqrt{\dfrac{400}{3}}$
D. $\sqrt{\dfrac{400}{4}}$
Let the body have a small angular displacement $\theta$ about the hinge point. \[\mathrm{Restoring\ torque} =\left( kr\theta \right) r =kr^{2}\theta,\\ \mathrm{Disturbing\ Torque}= I\alpha = I \ddot{\theta}\] \[I\ddot{\theta} =kr^{2}\theta \] \[I\ddot{\theta} +kr^{2}\theta =0\] \[\ddot{\theta} +\dfrac {kr^{2}}{I}\theta =0\] \[\omega_{n}=\sqrt {\dfrac {kr^{2}}{I}}\] \[I=m\left( 2r^{2}\right) =4mr^{2}\] \[\omega_{n} =\sqrt {\dfrac {kr^{2}}{4mr^{2}}} =\sqrt {\dfrac {k}{4m}} =\sqrt {\dfrac {400}{4}} \ \mathrm{rad/s}\]
Correct Answer: D
Solved Example: 38108
A single degree of freedom massspringviscous damper system with mass m, spring constant k and viscous damping coefficient q is critically damped. The correct relation among m, k and q is (GATE ME 2016 Shift 2)
A. q = $\sqrt{2km}$
B. q = 2$\sqrt{km}$
C. q = $\sqrt{\dfrac{2k}{m}}$
D. q = 2 $\sqrt{\dfrac{k}{m}}$
A critically damped system is defined as the system that comes to neutral (equilibrium) position as quickly as possible, without any overshoot.
Correct Answer: B
Solved Example: 38109
A static deflection of a spring under gravity, when a mass of 1 kg is suspended from it, is 1 mm. Assume the acceleration due to gravity g= 10 m/s$^2$. The natural frequency of this spring mass system (in rad/s) is: (GATE ME 2016 Shift 3)
A. 1 rad/s
B. 10 rad/s
C. 100 rad/s
D. 1000 rad/s
\[\omega_n = \sqrt{\dfrac{g}{\delta}} = \sqrt{\dfrac{10}{10^{3}}} = \sqrt{10^4} = 10^2 =100\ rad/s\]
Correct Answer: C
Solved Example: 38110
A mass is attached to two identical springs having spring constant k as shown in the figure. The natural frequency $\omega$ of this single degree freedom system is: (GATE ME 2017 Shift 2)
A. $\sqrt{\dfrac{2k}{m}}$
B. $\sqrt{\dfrac{k}{m}}$
C. $\sqrt{\dfrac{k}{2m}}$
D. $\sqrt{\dfrac{4k}{m}}$
For parallel springs, \[k_{eq} = k + k = 2k\] \[ \omega = \sqrt{\dfrac{2k}{m}}\]
Correct Answer: A
Torsional Vibrations
Learning Objectives:

Understand the significance of torsional vibrations in shafts.
Using the relation, net torque = moment of inertia . angular acc. \[\Sigma T = I \alpha = I \ddot{\theta}\] \[I \ddot{\theta} = k \theta\] negative sign is due to restoring nature of the force.
\[\ddot{\theta} + \frac{k_t}{I} \theta = 0\]
Angular velocity,
\[\omega_n = \sqrt{\dfrac{k_t}{I}}\]
Period of oscillation, \[\tau_n = \dfrac{2\pi}{\omega_n} = \dfrac{2\pi}{\sqrt{\dfrac{k_t}{I}}}\]
Solved Example: 38201
A solid disk with radius 'a' is connected to a spring at a point 'd' above the center of the disk. The other end of the spring is fixed to a vertical wall. The disk is free to roll without slipping on the ground. The mass of disk is 'M' and spring constant is 'K'. The polar moment of inertia for the disk about its center is J = Ma$^2$/2.
The natural frequency of this system in rad/s is given by: (GATE Mechanical 2016 Shift 1)
A. $\sqrt{\dfrac{2K(a+d)^2}{3Ma^2}}$
B. $\sqrt{\dfrac{2K}{3M}}$
C. $\sqrt{\dfrac{2K(a+d)^2}{Ma^2}}$
D. $\sqrt{\dfrac{K(a+d)^2}{Ma^2}}$
The total energy of the system remains constant.
Rotational K.E. of the disk= \(\dfrac {1}{2}I\dot{\theta}^{2}\)
Translational energy of centroid of disk = \(\dfrac {1}{2}Mv^{2}\) = \(\dfrac {1}{2}M(a \dot{\theta})^{2}\)
spring energy = \(\dfrac {1}{2}Kx^{2}\)
Cancelling $\dot{\theta}$,
\[\dfrac{3}{2}Ma^{2} \ddot {\theta }+K\left( a+d\right) ^{2}\theta=0\] \[\ddot {\theta }\left[ \dfrac {3Ma^{2}}{2}\right] +K\left( a+d\right) ^{2}\theta =0\] \[\omega_{n}=\sqrt {\dfrac {2K\left( a+d\right) ^{2}}{3Ma^{2}}}\]Correct Answer: A
Solved Example: 38202
The radius of gyration of a compound pendulum about the point of suspension is 100 mm. The distance between the point of suspension and the center of mass is 250 mm.
Considering the acceleration due to gravity as 9.81 m/s$^2$, the natural frequency of the compound pendulum, is: (GATE Mechanical 2017)
A. 12.43 rad/s
B. 15.66 rad/s
C. 17.81 rad/s
D. 19.22 rad/s
\begin{align*} \omega_n &= \sqrt{\dfrac{mgL}{I}}\\ &= \sqrt{\dfrac{mgL}{mk^2}}\\ &= \sqrt{\dfrac{9.81 \times 0.25}{(0.1)^2}}\\ &= 15.66\ \mathrm{rad/s} \end{align*}
Correct Answer: B
Forced Vibrations
Learning Objectives:

Model and solve systems involving harmonically forced vibrations.
Here, the system is excited with external forces, which most of the time periodic in nature. We can assume them as sine wave or cosine waves.
The equation of motion becomes,
\[m\ddot{x} + c\dot{x} + kx = F\] This is nonhomogeneous differential equation as the RHS is nonzero. The solution will have two components:

The homogeneous part, \(x_h\), which will eventually reduce to zero.

The particular part, \(x_p\), which will remain as a steady state solution.
Total solution = homogeneous part + particular part \[x = x_h +x_p\]
Case I: Undamped Forced Vibrations:
since c = 0, \[m\ddot{x} + kx = F(t)\] Assuming \[F(t) = F_0 \cos\omega t\] \[x_p = x \cos \omega t\] where, \[x = \dfrac{F_0}{km\omega^2}\] \[x = \dfrac{\delta_{st}}{1 r^2}\] where, \(r\) = frequency ratio = \(\dfrac{\omega}{\omega_n}\)
When r = 1, x \(\to\) \(\infty\), which is resonance.
Case II: Damped Forced Vibrations:
\[m\ddot{x} + c\dot{x} + kx = F_0 \cos\omega t\] \[x = \dfrac{\dfrac{F_0}{k}}{\sqrt{(1  r^2)^2 + {2\zeta r}^2}}\] \[\tan \phi = \dfrac{2\zeta r}{1  r^2}\] Magnification factor, \[M = \dfrac{x}{\delta_{st}} = \dfrac{1}{\sqrt{(1  r^2)^2 + {2\zeta r}^2}}\]
Solved Example: 38301
The equation $m \dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} + Kx = F_0 \sin \omega t$ is a second order differential equation. The solution of this linear equation is given as:
A. Complementary function
B. Particular function
C. Sum of complementary and particular function
D. Difference of complementary and particular function
The second order differential equation will always have complementary function in its answer. Also, since the RHS of the equation is nonzero, there will be particular integral too.
Correct Answer: C
Solved Example: 38302
What is meant by phase difference or phase angle in forced vibrations?
A. Difference between displacement vector (x$_p$) and velocity vector V$_p$
B. Angle in which displacement vector leads force vector by ($F_0 \sin \omega t$)
C. Angle in which displacement vector (x$_p$) lags force vector ($F_0 \sin \omega t$)
D. None of the above
Correct Answer: C
Solved Example: 38303
A singledegreefreedom springmass system is subjected to a sinusoidal force of 10N amplitude and frequency $\omega$ along the axis of the spring. The stiffness of the spring is 150 N/m, damping factor is 0.2 and the undamped natural frequency is 10$\omega$.
At steady state, the amplitude of vibration (in m) is approximately: (GATE ME 2015 Shift 2)
A. 0.05
B. 0.07
C. 0.70
D. 0.90
\[A=\dfrac {\dfrac {F_{0}}{k}}{\sqrt{\left[ 1\left( \dfrac {\omega}{\omega_{n}}\right) ^{2}\right] ^{2}+\left[ 2\xi \left( \dfrac {\omega }{\omega _{n}}\right) \right] ^{2}}}\] \[F_{0} =10N,\ k =150N/m,\ \dfrac {\omega}{\omega_{n}} =0.1,\ \zeta =0.2\] Substituting, \[A=0.066\approx 0.07 \ m\]
Correct Answer: B
Solved Example: 38304
The damping ratio for a viscously damped spring mass system, governed by the relationship \[m\dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} + kx = F(t)\] is given by: (GATE ME 2017 Shift I)
A. $\sqrt{\dfrac{c}{mk}}$
B. $\dfrac{c}{2\sqrt{km}}$
C. $\dfrac{c}{\sqrt{km}}$
D. $\sqrt{\dfrac{c}{2mk}}$
$\zeta = \dfrac{c}{c_c} = \dfrac{c}{2\sqrt{km}}$
Correct Answer: B
Solved Example: 38305
A machine of mass m = 200 kg is supported on two mounts, each of stiffness k = 10 kN/m. The machine is subjected to an external force (in N) $F(t) = 50 \cos(5t)$.
Assuming only vertical translatory motion, the magnitude of the dynamic force (in N) transmitted from each mount to the ground is: (GATE Mechanical 2018 Shift I)
A. 33.33 N
B. 39.90 N
C. 48.02 N
D. 89.22 N
F = $F_0 \sin \omega t$, F$_0$ = 50, $\omega$ = 5, m = 200 kg, no damping $\zeta$ = 0 \[\omega_n = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{20000}{200}} = 10\] \[\dfrac{\omega}{\omega_n} = \dfrac{5}{10} = 0.5\] \[k_{eq}= 2k = 2000 N/m^2\] \[\dfrac{F_t}{F_0} = \sqrt{\dfrac{1+ \left(2\zeta \dfrac{\omega}{\omega_n}\right)^2}{\left[1\left(\dfrac{\omega}{\omega_n}\right)^2\right]^2+ \left(2\zeta \dfrac{\omega}{\omega_n}\right)^2}} = \sqrt{\dfrac{1}{10.5^2}} = 1.25\] \[F_t = 1.25 \times F_0 = 1.25 \times 50 = 66.66\ N/m^2\] So, each mount will share half = 33.33 N force.
Correct Answer: A