Forced Vibration MCQ Quiz - Objective Question with Answer for Forced Vibration - Download Free PDF

Explanation:

In the context of system dynamics, both mechanical and electrical systems can be modeled to understand how they behave in response to inputs over time. These models often involve elements that have analogous properties in both domains, allowing engineers to draw parallels between them. For vibrating systems, the equivalences between mechanical and electrical systems are particularly useful for analysis and design.

Mechanical System Components:

• Mass (m): Represents the inertia of the system, resisting acceleration when a force is applied.
• Spring Stiffness (k): Represents the restoring force that acts to bring the system back to its equilibrium position.
• Damping Coefficient (c): Represents the resistive force that dissipates energy, reducing the amplitude of oscillations.
• Force (F): The external input or excitation applied to the system.

Electrical System Components:

• Inductance (L): Represents the inertia of the electrical system, resisting changes in current.
• Capacitance (C): Represents the ability to store electrical energy, analogous to the spring in a mechanical system.
• Resistance (R): Represents the resistive element that dissipates energy, analogous to the damping coefficient in a mechanical system.
• Voltage (V): The external input or excitation applied to the system.

Damping Coefficient (c) and Resistance (R):

• The damping coefficient in a mechanical system is a measure of the resistive force that opposes the motion and dissipates energy. This is analogous to resistance in an electrical system, which opposes the flow of current and dissipates electrical energy as heat. Both elements serve to reduce the amplitude of oscillations in their respective systems, thereby controlling the stability and response of the system.

Concept:

The damped natural frequency of a vibrating system is given by

\( \omega_d = \omega_n \sqrt{1 - \zeta^2} \), where \( \omega_n = \sqrt{\frac{k}{m}} \) is the natural frequency and \( \zeta = \frac{c}{2\sqrt{km}} \) is the damping ratio.

Given:

Mass, m = 200 kg

Spring stiffness, k = 80 N/mm = 80000 N/m

Damping coefficient, c = 800 Ns/m

Calculation:

Undamped natural frequency, \( \omega_n = \sqrt{\frac{80000}{200}} = \sqrt{400} = 20~\text{rad/s} \)

Damping ratio, \( \zeta = \frac{800}{2\sqrt{80000 \cdot 200}} = \frac{800}{2 \cdot 4000} = 0.1 \)

Damped natural frequency, \( \omega_d = 20 \cdot \sqrt{1 - 0.1^2} = 20 \cdot \sqrt{0.99} = 19.8998~\text{rad/s} \)

Damped frequency in Hz, \( f_d = \frac{\omega_d}{2\pi} = \frac{19.8998}{6.283} \approx 3.17~\text{Hz} \)

Now, \( \frac{3\sqrt{11}}{\pi} = \frac{3 \times 3.3166}{3.14} \approx \frac{9.9498}{3.14} \approx 3.17~\text{Hz} \)

Explanation:

First, let's convert the operational speed from rpm to radians per second:

Operational speed (ω) in radians per second = 2π × (rpm / 60)

Given that the rpm is 450, we have:

ω = 2 × π × (450 / 60)

ω = 2 × π × 7.5

ω = 15π rad/s

Given that π² = 10, we can find π by taking the square root of 10:

π = √10

Now, ω = 15 × √10 rad/s

Next, let's calculate the natural frequency (ωₙ) of the system using the formula:

ωₙ = √(k/m)

Where m is the mass of the refrigerator unit (44 kg) and k is the stiffness of each spring. Since there are 3 springs of the same stiffness, the effective stiffness (kₑ) is:

kₑ = 3k

Therefore, the formula for the natural frequency becomes:

ωₙ = √(3k / 44)

To determine the allowable stiffness, we use the given condition that only 10% of the shaking force is allowed to be transmitted to the supporting structure. This corresponds to a transmissibility ratio (T) of 0.1.

The transmissibility ratio (T) is given by:

\( T_r = \frac{1}{\sqrt{(1 - r^2)^2 + (2 \zeta r)^2}} \)

Where ζ is the damping ratio and r is the frequency ratio (ω/ωₙ).

For a system with negligible damping (ζ ≈ 0), the transmissibility ratio simplifies to:

T = 1 / √[1 + r²]

Given T = 0.1, we have:

0.1 = 1 / √[1 + r²]

Squaring both sides:

(0.1)² = 1 / (1 + r²)

0.01 = 1 / (1 + r²)

1 + r² = 100

r² = 99

r = √99

Substituting r = ω/ωₙ, we get:

√99 = (15 × √10) / ωₙ

ωₙ = (15 × √10) / √99

ωₙ = 15 × √(10/99)

ωₙ = 15 × √(10/99)

ωₙ = 15 × √(10/99)

ωₙ = 15 × (√10 / √99)

ωₙ = 15 × (√10 / √99)

ωₙ = 15 × (1 / √9.9)

ωₙ = 15 × (1 / 3.146)

ωₙ ≈ 4.77 rad/s

Using the formula for natural frequency:

ωₙ = √(3k / 44)

4.77 = √(3k / 44)

Squaring both sides:

(4.77)² = 3k / 44

22.7529 = 3k / 44

3k = 22.7529 × 44

3k = 1001.1276

k = 1001.1276 / 3

k ≈ 333.71 N/m

Converting to N/mm:

k ≈ 333.71 / 1000 N/mm

k ≈ 0.334 N/mm

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then 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 frequency ratio ω/ω_n > √2, then 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 ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping

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:

At resonance (ω = ω_n), phase angle ϕ is 90°.

When ω/ω_n ≫ 1; the phase angle is very close to 180°. Here inertia force increases very rapidly and its magnitude is very large.

Concept:

In vibration system, transmissibility is given by

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

Where F_T = maximum force transmitted to the ground

\(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi \omega }{{{\omega }_{n}}} \right)}^{2}}}}\)

Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

K = spring stiffness, m = mass

Calculation:

Given, Harmonic excitation force: F(t) = F₀cos(ωt)

F_T = F_O

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

∴ Transmissibility, ϵ = 1

Also, \(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{w}{{{w}_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{w}{{{w}_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi w}{{{w}_{n}}} \right)}^{2}}}}=1\)

Let \(\frac{\omega }{{{\omega }_{n}}}=x\)

So, 1 + (2ξ x)² = (1 – x²)² + (2ξ x)²

(1 – x²)² = 1

1 – x² = ± 1

Taking +ve sign x² = 0

∴ Taking –ve sign

x² = 2

Putting value of x:

(ω/ω_n)² = 2

\(\frac{\omega }{{{\omega }_{n}}}=\sqrt{2}\)

∴ ω = √2ω_n

∵ Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

Explanation:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < √2, then 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 frequency ratio ω/ωn > √2, then 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 ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then 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 frequency ratio ω/ω_n > √2, then 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 ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping

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,

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

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

Explanation:

Vibration Isolation: 

• The purpose of vibration isolation is to control the transmission of the vibration to the base upon which the machines are installed.
• It is done by mounting the machines on the spring, dampers, or other vibration isolation material. 
  • Force transmissibility is defined as the ratio of force transmitted to the foundation that impressed on the system.
  • For a viscous damped system with impressed force F0 and transmitted force FT, transmissibility is given as 

\(\frac{{{F_T}}}{{{F_0}}} = {\rm{\;Transmissibility\;}} = \frac{{\sqrt {1 + {{\left( {2\xi r} \right)}^2}} }}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Where, ω = speed of the exciting source, rad/s, ωn = natural frequency of system, rad/s, ζ = damping ratio

The transmissibility curve for different values of the damping ratio is shown below

Conclusions:

• Independent of the value of the damping ratio, the transmissibility of the mechanical system tends to zero as the value of the frequency ratio is above √2. The section beyond the frequency ratio \(\sqrt2\)  is known as the Isolation part of the transmissibility curve i.e., For effective vibration isolation, the frequency ratio \(\frac{\omega_n}{\omega}\) must be less than \(\frac{1}{\sqrt2}\)
• If the frequency ratio (ω/ωn) is less than\(\sqrt2\) then the transmitted force is always greater than the exciting force.
• If the frequency ratio (ω/ωn) is equal to \(\sqrt2\) then the transmitted force is equal to the exciting force.

Explanation:

Damping ratio:

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

\(\xi = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} = \frac{c}{{2m{\omega _n}}}\)

Transmissibility:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility.

\(\varepsilon = \frac{{\sqrt {1 + {{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2} + {{\left( {1 - \;\frac{{{\omega ^2}}}{{{\omega _n}^2}}} \right)}^2}\;} }}\)

If \(\frac{\omega}{{{\omega_n}}} = \sqrt 2 \;\) then ϵ = 1 for all values of damping factor c/cc.

Magnification factor:

The ratio of the amplitude of forced vibration by the harmonic force Focos(ωt) to the static deflection produced by force Fo is known as the magnification factor.

\(MF=\frac{A}{{{X_s}}} = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Concept:

Vibration Isolation: The purpose of vibration isolation is to control the transmission of the vibration to the base upon which the machines are installed. It is done by mounting the machines on the spring, dampers, or other vibration isolation material. 

• Force transmissibility is defined as the ratio of force transmitted to the foundation that impressed on the system.
• For a viscous damped system with impressed force F₀ and transmitted force F_T, transmissibility is given as 

\(\frac{{{F_T}}}{{{F_0}}} = {\rm{\;Transmissibility\;}} = \frac{{\sqrt {1 + {{\left( {2\xi r} \right)}^2}} }}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Where, ω = speed of the exciting source, rad/s, ω_n = natural frequency of system, rad/s, ζ = damping ratio

The transmissibility curve for different values of the damping ratio is shown below

Conclusions:

Independent of the value of the damping ratio, the transmissibility of the mechanical system tends to zero as the value of the frequency ratio is above √2. The section beyond frequency ratio √2 is known as the Isolation part of the transmissibility curve.

If the frequency ratio (ω/ω_n) is less than √2 then transmitted force is always greater than the exciting force.

If the frequency ratio (ω/ωn) is equal to √2 then transmitted force is equal to the exciting force.

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  

Concept:

Impulse (I) = Change in momentum (ΔP)

Calculation

Given

Mass, m = 5 kg

Stiffness of spring, k = 10 kN/m

At, t = 0 , Mass is at rest u = 0

Force, F = 5 kN is applied for time, t = 10^-4 sec

Using impulse momentum theorem

Impulse (I) = Change in momentum (ΔP)

F × t = m × (v – u)

5 × 10³ × 10^-4  = 1 × (v – 0)

\(v = 0.5~\frac{m}{{sec}}\)

Using energy balance

\(\frac{1}{2}m{v^2} = \frac{1}{2}k{X^2}\)

\(1 \times {0.5^2} = 10 \times {10^3} \times {X^2}\)

X = 5 × 10^-3 = 5 mm