### Free Vibrations

• Calculate the natural frequency vibration for an undamped free vibration.

• Understand what do you mean by damping and its effect in a spring-damper 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:

1. 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.

2. 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.

3. 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:

#### 38-1-01

In which type of vibrations, amplitude of vibration goes on decreasing every cycle? (Based on TANGEDCO AE ME 2015)

Solution:
Damping of vibrations results in dissipation of energy due to friction, noise or unwanted signals, thereby reduction in amplitude of oscillatory motion.

### Solved Example:

#### 38-1-02

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)

Solution:
The natural frequency $\omega_n = \sqrt{\dfrac{k}{m}}$ does not depend upon gravity.

### Solved Example:

#### 38-1-03

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)

Solution:
Damping ratio, \begin{align*} \xi &= \dfrac{C}{C_{c}} = \dfrac{25}{2\sqrt{1 * 100}} =\dfrac{25}{20} =1.25 \end{align*}

### Solved Example:

#### 38-1-04

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)

Solution:

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*}

### Solved Example:

#### 38-1-05

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:

Solution:
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: $\omega = \sqrt{\dfrac{k}{m}} = 52.54 Hz$

### Solved Example:

#### 38-1-06

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)

Solution:
$S=\dfrac {FL^{3}}{3EI}$ $k=\dfrac {F}{s}=\dfrac {3EI}{l^{3}} =\dfrac {3EI}{0.01^{3}} =3\times 10^{6}EI$ $w_{n} =\sqrt {\dfrac {k}{m}} =\sqrt {\dfrac {3\times 10^{6}EI}{5}}w_{n} =2449.48\sqrt {EI}$ \begin{align*} f_{n} &= \dfrac {\omega _{n}}{2\pi}\\ 100 &= \dfrac {2449.48\sqrt {EI}}{2\pi}\\ EI &= 0.065N.m^{2} \end{align*}

### Solved Example:

#### 38-1-07

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)

Solution:
Let the body have a small angular displacement $\theta$ about the hinge point. $\mathrm{Restoring\ torque} =-\left( kr\theta \right) r =-kr^{2}\theta,\\ \mathrm{Disturbing\ Torque}= I\alpha = I \ddot{\theta}$ $I\ddot{\theta} =-kr^{2}\theta$ $I\ddot{\theta} +kr^{2}\theta =0$ $\ddot{\theta} +\dfrac {kr^{2}}{I}\theta =0$ $\omega_{n}=\sqrt {\dfrac {kr^{2}}{I}}$ $I=m\left( 2r^{2}\right) =4mr^{2}$ $\omega_{n} =\sqrt {\dfrac {kr^{2}}{4mr^{2}}} =\sqrt {\dfrac {k}{4m}} =\sqrt {\dfrac {400}{4}} \ \mathrm{rad/s}$

### Solved Example:

#### 38-1-08

A single degree of freedom mass-spring-viscous 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)

Solution:

A critically damped system is defined as the system that comes to neutral (equilibrium) position as quickly as possible, without any overshoot.

### Solved Example:

#### 38-1-09

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)

Solution:
$\omega_n = \sqrt{\dfrac{g}{\delta}} = \sqrt{\dfrac{10}{10^{-3}}} = \sqrt{10^4} = 10^2 =100\ rad/s$

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)
For parallel springs, $k_{eq} = k + k = 2k$ $\omega = \sqrt{\dfrac{2k}{m}}$