Damping Coefficient and Damping Ratio MCQ Quiz - Objective Question with Answer for Damping Coefficient and Damping Ratio - Download Free PDF

Concept:

Damping ratio:

The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

\(\zeta = \frac{C}{{{C_c}}}=\frac{C}{2 \sqrt{km}}\)

Also, \(2 ξ ω_n=\frac Cm\)

• Overdamped System: ζ > 1
• Underdamped: ζ < 1
• Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion. 

Calculation:

Given:

C = 10 Ns/m, m = 5 kg, ω_n =  240 rad/min = 4 rad/sec

Therefore, for damping ratio, 

\(2 ξ ω_n=\frac Cm\)

\(\Rightarrow 2 ξ \times 4=\frac {10}{5}\)

ξ = 0.25

Concept:

The damping ratio (ξ) is defined as ratio of viscous damping coefficient to critical viscous damping coefficient.

\(⇒ {{ξ }} = \frac{{{c}}}{{2\sqrt {{{km}}} }}{{\;\;\;\;\;}} \ldots \left( 1 \right)\)

where, ξ = damping ratio, c = damping coefficient, k = spring constant, 

Calculation: 

Given: 

 m = 2 kg, k = 200 N/m, c = 10 Ns/m

Using equation (1),

\(⇒ {{ξ }} = \frac{{10}}{{2\sqrt {200 \times 2} }}{{}}=\frac{10}{40}=0.25\)

⇒ ξ = 0.25

Concept:

Damping ratio:

The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

\(\zeta = \frac{C}{{{C_c}}}=\frac{C}{2 \sqrt{km}}\)

Also, \(2 ξ ω_n=\frac Cm\)

• Overdamped System: ζ > 1
• Underdamped: ζ < 1
• Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion. 

Calculation:

Given:

C = 10 Ns/m, m = 5 kg, ω_n =  240 rad/min = 4 rad/sec

Therefore, for damping ratio, 

\(2 ξ ω_n=\frac Cm\)

\(\Rightarrow 2 ξ \times 4=\frac {10}{5}\)

ξ = 0.25

Explanation:

In forced vibration, the transmissibility ratio is given by,

\(\Rightarrow TR = \dfrac{{{F_T}}}{{{F_0}}} = \dfrac{{\sqrt {1 + {{\left( {2\xi q} \right)}^2}} }}{{\sqrt {{{\left( {1 - {q^2}} \right)}^2} + {{\left( {2\xi q} \right)}^2}} }}\)

Where,

\(q = {\bf{Frequency}}\;{\bf{ratio}} = \dfrac{\omega }{{{\omega _n}}}\)

• When, q = 0 ⇒ TR = 1, (independent of ζ)
• When, q = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When, q = √2, ⇒ All the curves pass through the point TR = 1 for all values of damping factor ξ.
• When q < √2 ⇒ TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When q > √2 ⇒ TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus, vibration isolation is possible only in the range of q > √2. Here the force transmitted to the foundation increases as the damping is increased.

Explanation:

• Damping is a force that opposes motion and removes energy from the system, usually due to friction or resistance.

• In Simple Harmonic Motion (SHM), when damping is introduced:

  • The amplitude of oscillation gradually reduces over time.

  • The system continues to oscillate, but with less energy in each cycle.

  • The frequency may slightly decrease depending on the damping level, but for small damping, frequency change is negligible.

Additional Information

• Undamped SHM: Constant amplitude, constant frequency.

• Damped SHM: Amplitude decays exponentially with time.

• Overdamping/Critical damping: Oscillation may stop altogether.

Concept:

\({\rm{Damping\;factor\;}} = \zeta = \frac{c}{{2\sqrt {km} }}\)

Where,

ζ = damping factor, c = coefficient of damping, k = spring constant, m = mass

Calculation:

Given:

\(\frac{{{d^2}x}}{{d{t^2}}} + 3\frac{{dx}}{{dt}} + 9x = 10\sin \left( {5t} \right)\)

By comparing this equation with \(m\ddot x + c\dot x + kx = F(t)\)

m = 1 kg, c = 3 N s/m, k = 9 N/m,

\({\rm{\zeta }} = \frac{3}{{2\;\sqrt {9 \times 1} \;}}\)

\({\rm{\zeta \;}} = \frac{3}{6}\)

Concept:

Steady state Amplitude:

\(A = \frac{{{f_0}/k}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\varepsilon \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

At resonance ω = ω_n i.e. damped frequency is equal to natural frequency

\(A = \frac{{{f_0}/k}}{{2\varepsilon }}\)

Calculation:

\({\left( {Amplitude} \right)_{Resonance}} = \frac{{{F_0}}}{{2\varepsilon k}} = \frac{{100}}{{2 \times 0.25 \times 10 \times 1000}}\)

x = 0.02 m = 20 mm  

Explanation:

Overdamped and critically damped systems return to their equilibrium position without any oscillation. Critically damped systems return faster than Overdamped systems.

Critical Damping (ζ = 1):

\(x(t) = (A + Bt){e^{ - {\omega _n}t}}\)

The displacement will be approaching zero with the shortest possible time. Due to this reason, large military field guns are critically damped.

Overdamped System (ζ > 1): 

\(x(t) = A{e^{( - \xi + \sqrt {{\xi ^2} - 1} ){ω _n}t}} + B{e^{( - \xi - \sqrt {{\xi ^2} - 1} ){ω _n}t}}\)

This is the equation of aperiodic motion i.e. the system cannot vibrate due to over-damping. The magnitude of the resultant displacement approaches zero with time.

Underdamped systems return to their equilibrium position with oscillation.

If the system is underdamped it will swing back and forth with decreasing size of the swing until it comes to a stop. Its amplitude will decrease exponentially.

\(x\left( t \right) = {e^{ - \xi {ω _n}t}}\left( {A{e^{i{ω _d}t}} + B{e^{ - i{ω _d}t}}} \right)\)

Explanation:

In forced vibration, the transmissibility ratio is given by,

\(\Rightarrow TR = \dfrac{{{F_T}}}{{{F_0}}} = \dfrac{{\sqrt {1 + {{\left( {2\xi q} \right)}^2}} }}{{\sqrt {{{\left( {1 - {q^2}} \right)}^2} + {{\left( {2\xi q} \right)}^2}} }}\)

Where,

\(q = {\bf{Frequency}}\;{\bf{ratio}} = \dfrac{\omega }{{{\omega _n}}}\)

• When, q = 0 ⇒ TR = 1, (independent of ζ)
• When, q = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When, q = √2, ⇒ All the curves pass through the point TR = 1 for all values of damping factor ξ.
• When q < √2 ⇒ TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When q > √2 ⇒ TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus, vibration isolation is possible only in the range of q > √2. Here the force transmitted to the foundation increases as the damping is increased.

Concept:

Equation of motion for a single degree of freedom system with one damper is given by

\(m\ddot x + c\dot x + kx = 0\)

Critical damping coefficient, c_c = 2ζmω_n

\({c_c} = 2ζ m\sqrt {\frac{k}{m}} = 2ζ\sqrt {km}\)

Calculation:

Given:

ζ = 1 (for critically damped)

The equation of motion for a single degree of freedom system is

4ẍ + 9ẋ + 16x = 0

Comparing it with the equation of motion

\(m\ddot x + c\dot x + kx = 0\)

m = 4 kg, c = 9 N/m/s, k = 16 N/m

\(c_c=2\zeta \sqrt {km} = 2\;\sqrt {16 \times 4} = 16~kg/s\)

Concept:

Ratio of successive amplitude is given by

\(\frac{{{{\bf{X}}_0}}}{{{{\bf{X}}_1}}} = \frac{{{{\bf{X}}_1}}}{{{{\bf{X}}_2}}} = \frac{{{{\bf{X}}_2}}}{{{{\bf{X}}_3}}} = \ldots \ldots \ldots \ldots \ldots \ldots = \frac{{{{\bf{X}}_{\bf{n}}}}}{{{{\bf{X}}_{{\bf{n}} + 1}}}} = {{\bf{e}}^{\bf{δ }}}\)

where δ = Logarithmic amplitude

Logarithmic decrement is given by

\({\bf{δ }} = \frac{1}{{\bf{n-1}}}{\bf{ln}}\left( {\frac{{{{\bf{X}}_{\bf{o}}}}}{{{{\bf{X}}_{\bf{n}}}}}} \right)=\frac{1}{{\bf{n}}}{\bf{ln}}\left( {\frac{{{{\bf{X}}_{\bf{o}}}}}{{{{\bf{X}}_{\bf{n+1}}}}}} \right)\)

Damping ratio(ζ) is given by

\(ζ = \frac{δ }{{\sqrt {4{\pi ^2}\; + \;{δ ^2}} }}\)

Calculation:

Given:

X₀ = 0.7, X₁ = 0.28, X₂ = 0.25, X₃ = 0.23, X₄ = 0.067, ζ = ?

Now, we know that

\({\bf{δ }} = \frac{1}{{\bf{n}}}{\bf{ln}}\left( {\frac{{{{\bf{X}}_{\bf{o}}}}}{{{{\bf{X}}_{\bf{n}}}}}} \right) = \frac{1}{4}ln\left( {\frac{{0.7}}{{0.067}}} \right)\)

∴ δ = 0.587

The damping ratio is given by

\(ζ = \frac{\delta }{{\sqrt {4{\pi ^2}\; + \;{\delta ^2}} }} = \;\frac{{0.587}}{{\sqrt {4{\pi ^2}\; + \;{{0.587}^2}} }}\)

∴ ζ = 0.0928

Concept:

The logarithmic decrement of  viscous damped vibration is given as, \(δ =n{\rm{ξ \;}}{{\rm{ω }}_{{\rm{n\;}}}}{{\rm{T}}_{\rm{d}}}\)

where, ω_n = natural frequency and we can write \(\omega _n=\frac{\omega _d}{\sqrt{1-ξ ^2}}\) and \(T_d=\frac{2\pi}{\omega _d}\)

Calculation:

Given:

n = 5, δ = 8.11

Therefore, \(8.11 = 5 \times ξ \times \frac{{{\omega _d}}}{{\sqrt {1 - {ξ ^2}} }} \times \frac{{2\pi }}{{{\omega _d}}}\)

∴ ξ = 0.25 = 25%

Concept:

Then damped natural frequency is given by \({\omega _d} = \left( {\sqrt {1 - {\zeta ^2}} } \right){\omega _n}\)

Where ζ is the damping factor and ω_n is the natural frequency of the system

Natural Frequency \({\omega _n} = \sqrt {\frac{k}{m}} \)

damping factor \(\zeta = \frac{C}{{{C_c}}} = \frac{C}{{2\sqrt {km} }}\) 

Where C is the damping constant, k is the stiffness of the spring and m is mass of the system

Calculation:

Given, k = 3.6 kN/m = 3.6 ×10³ N/m, C = 400 Ns/m, m = 50 kg

Natural frequency \({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{3.6 \times {{10}^3}}}{{50}}} = 8.485\;rad/s\) 

and damping factor \(ζ = \frac{C}{{{C_c}}} = \frac{C}{{2\sqrt {km} }} = \frac{{400}}{{2\sqrt {3.6 \times {{10}^3} \times 50} }} = 0.471\)

Then damped natural frequency \({\omega _d} = \left( {\sqrt {1 - {ζ ^2}} } \right){\omega _n} = \left( {\sqrt {1 - {{0.471}^2}} } \right)8.485 = 7.48\)

\({f_d} = \frac{{{\omega _d}}}{{2\pi }} = \frac{{7.48}}{{2\pi }} = 1.19\;Hz\)

Explanation:

Damped vibration:

When the energy of a vibrating system is gradually dissipated by friction and other resistance, the vibrations are said to be damped vibration.

Damping ratio:

The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

\(\zeta = \frac{C}{{{C_c}}}\)

• Overdamped System: ζ > 1
• Underdamped: ζ < 1
• Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion. 

Concept:

The coefficient of damping is given as,

C = 2mζω_n where, C = coefficient of damping, m = mass of a body, ζ = damping factor, ω_n = natural frequency    

And Critical coefficient of damping is given as, C_c = 2mω_n  .......(ζ = 1 for Critical damping) 

Calculation:

Given:

m = 100 kg, ω_n = 10 rad/sec,

Therefore, C_c = 2 × 100 × 10 = 2000 N/m/sec