The document summarizes the analysis of forced vibrations in a single degree of freedom system. It presents the derivation of the steady state response using complex notation and defines the transfer function. It describes how the response amplitude and phase angle vary with frequency and damping ratio. It also defines concepts like magnification factor, quality factor, bandwidth and their relation to damping in the system.
Filters for Electromagnetic Compatibility Applications
ME421-SDF (Forced) part 2.pdf
1. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
Alternative derivation using complex notations
If forcing function 𝑓 𝑡 = 𝐹𝑒𝑖𝜔𝑡 then equation of motion is
𝑚 ሷ
𝑥 + 𝑐 ሶ
𝑥 + 𝑘𝑥 = 𝐹𝑒𝑖𝜔𝑡
Steady state response is of the form
𝑥𝑝 𝑡 = 𝑋𝑒𝑖(𝜔𝑡−𝜙)
= 𝑋𝑒𝑖𝜔𝑡. 𝑒−𝑖𝜙
= 𝑋𝑒−𝑖𝜙. 𝑒𝑖𝜔𝑡
= 𝑋∗𝑒𝑖𝜔𝑡
Where 𝑋∗ is a phasor quantity, i.e. a complex number having amplitude X and phase 𝜙
Eq. 21
Eq. 22
Differentiating Eq. 22 twice
ሶ
𝑥 𝑡 = 𝑋∗𝑖𝜔𝑒𝑖𝜔𝑡
ሷ
𝑥 𝑡 = −𝑋∗𝜔2𝑒𝑖𝜔𝑡
Now substituting in Eq. 21
−𝑚𝑋∗𝜔2𝑒𝑖𝜔𝑡 + 𝑐𝑋∗𝑖𝜔𝑒𝑖𝜔𝑡 + 𝑘𝑋∗𝑒𝑖𝜔𝑡 = 𝐹𝑒𝑖𝜔𝑡
−𝑚𝑋∗𝜔2 + 𝑐𝑋∗𝑖𝜔 + 𝑘𝑋∗ = 𝐹
By equating coefficients
2. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
By re-arranging equation
𝑋∗ =
𝐹
𝑘 − 𝑚𝜔2 + 𝑖𝜔𝑐
Divide numerator and denominator by k
𝑋 =
𝐹
𝑘
1 −
𝑚𝜔2
𝑘
2
+
𝜔𝑐
𝑘
2
𝑋 = 𝑋∗ =
𝐹
(𝑘 − 𝑚𝜔2)2+(𝜔𝑐)2
1
2
1
𝑘 − 𝑚𝜔2 + 𝑖𝜔𝑐
= 𝐻 𝑖𝜔 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑
𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝐹𝑅𝐹
𝑋 =
𝐹
𝑘
1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
3. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
Above equation can be written as
𝑀 =
𝑋
𝛿𝑠𝑡
=
1
1 − 𝑟2 2 + 2𝜁𝑟 2
𝑤ℎ𝑒𝑟𝑒
𝐹
𝑘
= 𝛿𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑐 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑢𝑛𝑑𝑒𝑟 𝑠𝑡𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒
𝑋
𝛿𝑠𝑡
= M is called magnification factor , dynamic magnifier or amplitude ratio
𝜔
𝜔𝑛
= r, frequency ratio
𝜔, 𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝜔𝑛, natural frequency
𝜙 = 𝑡𝑎𝑛−1
2𝜁𝑟
1 − 𝑟2
Eq. 23 Eq. 24
General solution:
𝑥 𝑡 = 𝑥𝑐 𝑡 + 𝑥𝑝(𝑡)
= 𝑋0𝑒−𝜁𝜔𝑛𝑡𝑠𝑖𝑛 1 − 𝜁2 𝜔𝑛𝑡 + 𝜙 + 𝑋𝑠𝑖𝑛(𝜔𝑡 − 𝜙)
From Eq. 8A and Eq. 15
5. Don’t share on the web
In terms of phase angle
Single Degree of freedom system – Forced Vibrations
Observations:
• 𝑤ℎ𝑒𝑛 𝜁 = 0 𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜 , 𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛
𝑎𝑛𝑑 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝑓𝑜𝑟 0 < 𝑟 < 1
• 𝑤ℎ𝑒𝑛 𝜁 = 0 𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 180 , 𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛
𝑎𝑛𝑑 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑜𝑢𝑡 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒 𝑓𝑜𝑟 𝑟 > 1
• For 𝜁 > 0 and 0 < r < 1, the phase angle is
given by 0 < Φ < 90°, implying that the
response lags the excitation.
• For 𝜁 > 0 and r > 1, the phase angle is given by
90° < Φ < 180°, implying that the response
leads the excitation.
• 𝑤ℎ𝑒𝑛 𝑟 = 1, 𝜔 = 𝜔𝑛, 90𝜊 𝑝ℎ𝑎𝑠𝑒 𝑙𝑎𝑔 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑛𝑑 𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒
• 𝐹𝑜𝑟 𝑙𝑎𝑟𝑔𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑟 𝑡ℎ𝑒𝑟𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 180𝜊 𝑝ℎ𝑎𝑠𝑒 𝑙𝑎𝑔. 𝑅𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑎𝑛𝑑
𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑢𝑡 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒
6. Resonance, Damping, Bandwidth
Single Degree of freedom system – Forced Vibrations
How damped is it?
Damping in a system can be determined by noting the
maximum response, i.e., the response at the resonance
frequency. The max. value of the amplitude ratio at
resonance is called Q factor or quality factor. Sometimes
used in electrical engineering terminology, such as the
tuning circuit of a radio, where the interest lies in an
amplitude at resonance that is as large as possible.
The damping in a system is also indicated by the sharpness or width of the response curve
in the vicinity of a resonance frequency ωn
But in mechanical vibrations quality factor is defined as 𝑄 =
1
2𝜁
The Points R1 and R2 where amplification factor falls to
𝑄
2
are called half power points
because power absorbed by the damper. The difference between the frequencies associated
with R1 and R2 is called bandwidth of the system.
R1
R2
8. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
• Since 𝑅 > 0, then the range of values of 𝜁 for which Eq. 26 is valid is given by
1 − 2𝜁2 > 0 ⇒ 𝜁 <
1
2
Substituting Eq. 26 in Eq. 23 gives
𝑅 = 𝑀𝑚𝑎𝑥 =
1
2𝜁 1 − 𝜁2
• Now for 𝜁 ≤ 0.1 𝑀𝑚𝑎𝑥 occurs when 𝑅 ≈ 1
Substituting 𝑅 = 𝑟 ≈ 1 in Eq. 23 gives
𝑀𝑚𝑎𝑥 =
1
2𝜁
𝑓𝑜𝑟 𝜁 ≤ 0.1
The damping in a system is indicated by the sharpness of its response curve near resonance; it
Can be evaluated by the bandwidth of its half-power points
Eq. 27
9. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
Now at 𝑅1𝑎𝑛𝑑 𝑅2
𝑅1 = 𝑅2 =
𝑀𝑚𝑎𝑥
2
=
𝑄
2
Eq. 28
Substituting Eq. 27 and 28 in Eq. 23 gives
1
2𝜁
1
2
=
1
1 − 𝑟2 2 + 2𝜁𝑟 2
Squaring both sides simplifying gives
1 − 𝑟2 2 + 4𝜁2𝛽2 − 8𝜁2 = 0
Roots of this equation are
𝑟1,2 ≈ 1 ± 2𝜁
1
2 ≈ 1 ± 𝜁 𝑓𝑜𝑟 𝜁 ≤ 0.1
Therefore bandwidth = 𝑟2 − 𝑟1 = 1 + 𝜁 − (1 − 𝜁)
⇒
𝜔2 − 𝜔1
𝜔𝑛
= 2𝜁
𝐻𝑒𝑛𝑐𝑒 𝜁 =
𝜔2 − 𝜔1
2𝜔𝑛
Amplitude ratio at resonance
is called quality factor
R1 and R2 where M falls
to
𝑄
2
, called half power
points
Eq. 29
10. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
Frequency response function, FRF
11. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
12. Don’t share on the web
Single Degree of freedom system – Forced Vibrations
(a) (b)
(c)
Q1. Use the free body diagram to drive the equation of motion
Q2. A weight of 50 N is suspended from a
spring of stiffness 4000 N/m and is subjected to
a harmonic force of amplitude 60 N and
frequency 6 Hz. Find (a) the extension of the
spring due to the suspended weight, (b) the
static displacement of the spring due to the
maximum applied force, and (c) the amplitude
of forced motion of the weight.
Q3. Figure shows an automobile trailer which moves
over the road surface making approximately sinusoidal
profile. Trailer has a velocity of 60 Km/hr.
Calculate critical speed of the trailer if the vibration
amplitude is 1.5cm. Trailer mass is 50 kg. 𝛿𝑠𝑡 = 6𝑐𝑚
?
