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LEARNING OBJECTIVES

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 Introduction to undamped free vibrationIntroduction to undamped free vibration

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 Terminologies used in undamped free vibrationTerminologies used in undamped free vibration

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 Three methods to solve the Three methods to solve the undamped free vibration problemsundamped free vibration problems

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 Problems related to undamped free Problems related to undamped free vibration problems.vibration problems.

Free vibrations are oscillations about a systems equilibrium position that occur in the Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force. If during vibrations there is no loss of energy, absence of an external excitation force. If during vibrations there is no loss of energy, it is known as undamped vibration. The first step in solving a vibration problem is it is known as undamped vibration. The first step in solving a vibration problem is setting up the differential equation of motion

setting up the differential equation of motion

Consider spring mass system which is assumed to move only along the vertical Consider spring mass system which is assumed to move only along the vertical direction as shown below

direction as shown below. Let m be the mass of . Let m be the mass of the block and k be the stiffness the block and k be the stiffness of of the spring. When block of mass m is attached to spring , the deflection of spring will the spring. When block of mass m is attached to spring , the deflection of spring will be

be ∆∆ , known as static deflection. In the static equilibrium position, the free body, known as static deflection. In the static equilibrium position, the free body diagram of forces acting on the mass is shown in Figure(b). Hence mg= kA

diagram of forces acting on the mass is shown in Figure(b). Hence mg= kA Once the system is disturbed,

Once the system is disturbed, the system executes vibrations.the system executes vibrations.

Let at any instant of time t, the mass is displaced from the equilibrium position x, the Let at any instant of time t, the mass is displaced from the equilibrium position x, the different forces acting on the system are shown in figure (d)

different forces acting on the system are shown in figure (d) From Newton’s second law of motion

From Newton’s second law of motion

∑

^F^F ⁼^{= ma}^ma

Inertia force Inertia force

( disturbing force) = restoring force ( disturbing force) = restoring force

mg mg x

x k

k x

x m

&& &&

== −− ((∆∆ ++ )) ++

0 0 ))

(( == +

+ k k xx x

x m m

&& &&

or

++ (( xx )) == 00 m

m k x k

&& &&

equation 2 is the differential equation of motion for spring mass system shown in equation 2 is the differential equation of motion for spring mass system shown in figure. Comparing equation (2) with the equation of SHM

figure. Comparing equation (2) with the equation of SHM x x

&& &&

++ ω ω ²² (( xx )) == 00

since the vibrations of the above system are free( without the resistance of external since the vibrations of the above system are free( without the resistance of external forces) we can write

forces) we can write

sec sec //

rad m rad m

k k

n n == ω ω

time period , from the equation(1) mg = k∆

from the equation(1) mg = k∆

∆

Difference between the translation ( rectilinear) and rotational system of vibration.

Translatory Rotational

In the analysis the disturbing and restoring In the analysis the disturbing and restoring FORCES are considered

FORCES are considered

In the analysis In the analysis

In the analysis MASS Moment of Inertia In the analysis MASS Moment of Inertia (J) is considered

(J) is considered Linear stiffness K , in N/m is

Linear stiffness K , in N/m is consideredconsidered

Problems Problems

1.A mass of 10kg when suspended from a spring causes a static deflection of 1cm . 1.A mass of 10kg when suspended from a spring causes a static deflection of 1cm . Find the natural frequency of

Find the natural frequency of system.system.

2. A spring mass

2. A spring mass system has a spring system has a spring stiffness stiffness K N/m and a mass oK N/m and a mass of m Kg. It has af m Kg. It has a natural frequency of vibration 12 Hz. An extra 2kg mass coupled to it. then the natural frequency of vibration 12 Hz. An extra 2kg mass coupled to it. then the natural frequency reduces by 2 Hz. find K and m.

natural frequency reduces by 2 Hz. find K and m.

3. A steel wire of 2mm diameter and 30m long. It is fixed at the upper end and 3. A steel wire of 2mm diameter and 30m long. It is fixed at the upper end and carries a mass of m kg at its free end. Find m so that the frequency of longitudinal carries a mass of m kg at its free end. Find m so that the frequency of longitudinal vibration is 4 Hz.

vibration is 4 Hz.

4. A spring mass system has a natural period of 0.2 seconds. What will be the new 4. A spring mass system has a natural period of 0.2 seconds. What will be the new period, if the spring constant is 1) increased by 50% 2) decreased by 50%,

period, if the spring constant is 1) increased by 50% 2) decreased by 50%,

5. A spring mass system has a natural frequency of 10 Hz when the spring constant 5. A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 N/m, the frequency is altered by 45%, Find the mass and spring is reduced by 800 N/m, the frequency is altered by 45%, Find the mass and spring constant of the original system.

constant of the original system.

6. Determine the natural frequency of

6. Determine the natural frequency of system shown in fig is which system shown in fig is which shaft is supportedshaft is supported in SHORT bearings.

in SHORT bearings.

7 7 . Determine the natural frequency of system shown in fig is which shaft . Determine the natural frequency of system shown in fig is which shaft is supportedis supported in LONG bearings.

in LONG bearings.

where l is the length of bearing and E –

where l is the length of bearing and E – young’s modulus and I is moment of Inertia.young’s modulus and I is moment of Inertia.

7. Determine the natural frequency of

7. Determine the natural frequency of system shown in fig is which system shown in fig is which shaft is supportedshaft is supported in LONG bearings.

in LONG bearings.

8. A light cantilever beam of rectangular section( 5 cm deep x 2.5cm 8. A light cantilever beam of rectangular section( 5 cm deep x 2.5cm wide) has a mass fixed at its free end. Find the ratio of frequency of wide) has a mass fixed at its free end. Find the ratio of frequency of free lateral vibration in vertical plane to that

free lateral vibration in vertical plane to that in horizontal.in horizontal.

9.. Determine the natural frequency of simple pendulum 9.. Determine the natural frequency of simple pendulum

10. Homogeneous square plate of size l and mass m is suspended 10. Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure, Find the from the mid point of one of its sides as shown in figure, Find the frequency of vibration.

frequency of vibration.

11. A compound pendulum which is rigid body of mass m and it is pivoted at O. The 11. A compound pendulum which is rigid body of mass m and it is pivoted at O. The point of pivot is at distance d from the centre of gravity. It is free to rotate about its point of pivot is at distance d from the centre of gravity. It is free to rotate about its axis. Find the f

axis. Find the frequency of oscillation of such pendulum.requency of oscillation of such pendulum.

12. A connecting rod shown in fig is supported at the wrist pin end. It is displaced 12. A connecting rod shown in fig is supported at the wrist pin end. It is displaced and allowed to oscillate. The mass of rod is 5kg and centre of gravity is 20 cm from and allowed to oscillate. The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O. If the frequency of oscillation is 40 cycles/minute, calculate the the pivot point O. If the frequency of oscillation is 40 cycles/minute, calculate the mass moment of inertia about its C.G.

mass moment of inertia about its C.G.

13. A semi circular homogenous disc of radius r 13. A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as and mass m is pivoted freely about its centre as shown in figure. Determine the natural frequency shown in figure. Determine the natural frequency of oscillation.

of oscillation.

14.A simply supported beam of square cross 14.A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass section 5mmx5mm and length 1m carrying a mass of 0.575 kg at the

of 0.575 kg at the middle is found to have natural fmiddle is found to have natural f requency of 30 rad/sec. Determinerequency of 30 rad/sec. Determine young’s modulus of elasticity of

young’s modulus of elasticity of beam.beam.

15. A spring mass system, k1 and m have a natural frequency f1. Determine the 15. A spring mass system, k1 and m have a natural frequency f1. Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 2/3 f1.

lowers the natural frequency to 2/3 f1.

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM COMPLETE SOLUTION OF SYSTEM EXECUTING SHM The equation of motion of

The equation of motion of system executing SHM can be represented bysystem executing SHM can be represented by 0

&& &&

---(1)---(1)

0

The general solution of equation (1)

The general solution of equation (1) can be expressed ascan be expressed as ---(2) ---(2)

Where A and B are arbitrary constant which can be determined from the initial Where A and B are arbitrary constant which can be determined from the initial conditions of the system. Two initial conditions are to be specified to evaluate these conditions of the system. Two initial conditions are to be specified to evaluate these constants. x=x

constants. x=x00at t=0 andat t=0 and

x x &&

^{= V}^{= V}^o^o at t=0. substituting in the equation (2)at t=0. substituting in the equation (2)

Energy method energy. For free vibration of undamped system at any instant of time is partly kinetic energy. For free vibration of undamped system at any instant of time is partly kinetic and partly potential. The kinetic energy T is stored in

and partly potential. The kinetic energy T is stored in the mass by virtue of its the mass by virtue of its velocityvelocity where as the potential energy U is stored in the form of strain energy in elastic where as the potential energy U is stored in the form of strain energy in elastic deformation or work done in a force field such as gravity.

deformation or work done in a force field such as gravity.

The total energy being constant T+U = constant. Its rate of change The total energy being constant T+U = constant. Its rate of change

t)

=

Is the required completeIs the required complete solution

solution

k ll

k k

Is given by

[ [ ^T ^T ^U ^U ]] ⁰ ⁰

dt dt d

d + + = =

From this we get a differential equation of motion as well as

From this we get a differential equation of motion as well as natural frequency of thenatural frequency of the system.

system.

Determine the natural frequency of spring

Determine the natural frequency of spring mass system using energy method.mass system using energy method.

Determine the natural frequency of system shown in figure.

Determine the natural frequency of the system shown in figure. Is there any limitation Determine the natural frequency of the system shown in figure. Is there any limitation on the value of K. Discuss?

on the value of K. Discuss?

Determine the natural frequency of system shown below. Neglect the mass of ball.

m m

ll k

m m a a

θ θ

m m

θ θ

m

A string shown in figure is under tension T whic

A string shown in figure is under tension T which can be assumed to remain consh can be assumed to remain constanttant for small displacements. Find the natural frequency of vertical vibrations of spring.

for small displacements. Find the natural frequency of vertical vibrations of spring.

An acrobat acrobat 120kg 120kg walks walks on on a a tight tight rope rope as as shown shown in in figure. figure. The The frequency frequency of of vibration in the given position is vertical direction is 30 rad/s. Find the tension in the vibration in the given position is vertical direction is 30 rad/s. Find the tension in the rope.

rope.

A manometer has a uniform bore of cross sec

A manometer has a uniform bore of cross section area A. If the column of liquid of tion area A. If the column of liquid of length L and Density ρ is set into motion as shown in figure. Find t

length L and Density ρ is set into motion as shown in figure. Find t he frequency of he frequency of the resulting oscillation.

the resulting oscillation.

m m

ll aa

T T T

T

36m 36m 8m

k k

R R

k k

m m

r r

Find the expression for natural frequency of system shown in the figure. Neglect the Find the expression for natural frequency of system shown in the figure. Neglect the mass of the cantilever beam. Study the special case i) k=Infinity ii) I

mass of the cantilever beam. Study the special case i) k=Infinity ii) I = infinity.= infinity.

Determine the expression for the natural frequency of a system shown in figure. The Determine the expression for the natural frequency of a system shown in figure. The two discs are keyed to a common shaft and have a combined mass moment of two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O. It is assumed that the card attached to mass m inertia about centre of oscillation O. It is assumed that the card attached to mass m does not stretch and is

does not stretch and is always under tension.always under tension.

Determine the natural frequency Determine the natural frequency of a system shown

of a system shown

m m ll

Determine the expression for the natural frequency of the system shown in figure.

Assume that the wires connecting the mas

Assume that the wires connecting the masses do not stretch and are alwayses do not stretch and are always ins in tension.

tension.

Determine the natural frequency of spring mass system taking the MASS OF Determine the natural frequency of spring mass system taking the MASS OF SPRING (m

SPRING (mss ) into account.) into account.

M M₃₃ M

M₁₁

M2 M2

k2 k1 k2

k1

m m

k

In document Vibrations (Page 23-33)

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