### 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.

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

### Solved Example:

#### 38-1-10

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