Vibrations & Damping in Mechanical Engineering: Problem Set

This assignment solution provides detailed calculations and explanations for mechanical vibration problems. It covers topics such as angular velocity, time period, maximum acceleration and velocity, velocity and acceleration at specific distances from the center, and kinetic energy calculations. Furthermore, it delves into the concept of vibration as a periodic conversion of elastic and kinetic energy, explaining undamped natural frequency and deriving the equation for a damped forced system. The solution explores different damping scenarios (no damping, underdamped, critically damped, and overdamped) and their impact on system behavior. Finally, it addresses forced vibration, discussing the system's response to external forces, resonance conditions, and the relationship between forced frequency and natural frequency. The amplitude and phase angle for a specific forced vibration problem are also calculated. Desklib offers a wealth of similar solved assignments and past papers to aid students in their studies.

1. Angular velocity (ω)
ω=2 πf
ω=2 π × 10=20 π rad /s
Time period of one cycle ( T )
T = 1
f
T = 1
10 =0.1 s
2. Maximum acceleration
amax=ω2 A
amax= (20 π )2 × 180
1000 =72 π2 m/s2Maximum velocity
v=ωA
v=20 π × 180
1000 =3.6 π m/s
3. Velocity at 100 mm from centre
v=± ω √ A2−x2
v=± 20 π √ ( 180
1000 )
2
− ( 100
1000 )
2
=± 2.99 π m/s
4. Acceleration at 50 mm from centre
a=−ω2 x
a=− ( 20 π )2
( 50
1000 )=−20 π2 m/s2
5. Kinetic energy at 150 mm from centre
KE= 1
2 m ω2 ( A2−x2 )
KE= 1
2 ( 30
1000 ) ( 20 π ) 2
( 180
1000
2
− 150
1000
2
) =5.94 π2 × 10−2 J

2.1. Vibration is defined as periodic and cyclic conversion of elastic energy to kinetic energy
and vice-versa. A static system like bridge or skyscraper when acted upon by external
impulsive force has build-up of elastic energy by virtue of deflection and deformation of its
structural elements. This elastic energy is spent in acceleration the mass of the structure upon
removal of force which causes a back and forth motion about its undisturbed position. The
frequency of back and forth motion in absence of external force and damping is called un-
damped natural frequency of vibration.
2.2. The equation of damped forced system is given as
d2 x
d t2 +2ζ ωn
dx
dt + ωn
2 x=0
Where, ζ = atual damping
critical damping = c
cc
And critical damping coefficient cc=2 √km
The solution is
x (t )=C est
Where C and s are complex constants with s=−ωn (ζ ± i √1−ζ2 )
The four cases that arise are:
a. No damping, ζ =0
In this case is similar to un-damped harmonic oscillator and solution is x (t )=ei ωn t
b. Underdamped, 0< ζ <1
The solution is x (t )= ( C1 +C2 ) e−ωn ζt ( e√1−ζ 2
+e− √1−ζ 2
)i ωn t
The ( e√1−ζ 2
+e− √1−ζ 2
)i ωnt term is oscillatory combined with decaying exponential term
eωn ζt. The vibrations will die out sinusoidally.
c. Critically damped, ζ =1
There is only decaying exponential term ( C1+C2 ) e−ωn ζt. The system’s amplitude
exponentially decays to zero.
d. Overdamped, ζ >1
The solution is x ( t ) = ( C1 +C2 ) e−ωn ζt . There is no oscillations and oscillations decays
faster than critically damped case.
2.3. The ODE of forced vibration is
m ¨x + c ˙x+ kx=F0 sin2 πft
The steady state solution is
x ( t ) = X sin(2 πft + ϕ)
a. At forced vibration, the system vibrates at the frequency of force and the
amplitude is small.
b. The amplitude of resonant vibration depends upon difference between forced
frequency and natural frequency of system.

c. When the forced frequency is same as natural frequency then the body starts to
vibrate with high amplitudes.
d. The rate of energy transfer is very large and frequency of vibration lags or leads
force frequency by π
2 .
2.4. ζ = c
√kmζ = 150
√10000 ×5 =0.67f n= 1
2 π √ k
m
f n= 1
2 π √ 10000
5 =7.11 s−1
r = f
f n
= 30
7.11 =4.22
Amplitude is given by
X = F0
k
1
√ ( 1−r2 )2
+ ( 2 ζr )2
X = 400
10000
1
√ ( 1−4.222 )
2
+ ( 2× 0.67 × 4.22 ) 2
=2.26 × 10−3 m
Phase angle is given by
ϕ =arctan ( −2 ζr
1−r2 )ϕ =arctan ( −2 ×0.67 × 4.22
1−4.222 ) =0.32 rad