Mechanical Vibrations MCQ Quiz - Objective Question with Answer for Mechanical Vibrations - Download Free PDF

Concept:

The circular frequency (natural frequency in rad/s) for undamped free vibrations is given by:

\( \omega_n = \sqrt{\frac{k}{m}} \)

Where,

• \( k \) = stiffness of the spring (in N/m)
• \( m \) = mass (in kg)

Given:

Mass, \( m = 200~\text{kg} \)

Stiffness, \( k = 80~\text{N/mm} = 80 \times 10^3~\text{N/m} \)

Calculation:

\( \omega_n = \sqrt{\frac{80 \times 10^3}{200}} = \sqrt{400} = 20~\text{rad/s} \)

Concept:

When a mechanical system is subjected to harmonic excitation, the response of the system is influenced by the excitation frequency and the damping present.

The ratio of the amplitude of the system's response to the static deflection under the same force is called the Dynamic Magnification Factor (DMF).

General Formula:

\( \text{DMF} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}} \)

Where,
\( r = \frac{\omega}{\omega_n} \) is the frequency ratio
\( \xi \) is the damping ratio.

At Resonance:

At resonance, excitation frequency equals natural frequency, i.e., \( \omega = \omega_n \Rightarrow r = 1 \).

Substitute in the equation:

\( \text{DMF}_{\text{resonance}} = \frac{1}{\sqrt{(1 - 1)^2 + (2\xi)^2}} = \frac{1}{2\xi} \)

Explanation:

Dunkerley's Empirical Formula for Natural Frequency of Transverse Vibration:

• Dunkerley's empirical formula is a crucial method used to determine the natural frequency of transverse vibration for a shaft carrying a number of point loads and uniformly distributed load (UDL). This formula is particularly useful in mechanical and structural engineering applications where predicting the vibration characteristics of shafts is essential for ensuring stability and preventing resonance.
• The natural frequency of a system is the frequency at which it tends to oscillate in the absence of any driving or damping force. Dunkerley's formula provides a simplified approach to estimate the natural frequency of a complex system by considering the contributions of individual components separately.

Given:

Let \( f_n \) = Natural frequency of the complete system

\( f_{n1}, f_{n2}, f_{n3}, \dots, f_{ns} \) = Natural frequencies due to individual point loads and UDL

Calculation:

According to Dunkerley's empirical formula, the natural frequency is given by:

\( \frac{1}{f_n^2} = \frac{1}{f_{n1}^2} + \frac{1}{f_{n2}^2} + \frac{1}{f_{n3}^2} + \cdots + \frac{1}{f_{ns}^2} \)

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:

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

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

k_eq = 0.4 k

Concept:

\(Time~ Period= T =2\pi √{\frac Lg}\)

where L = Length of the pendulum, g = acceleration due to gravity

Calculation:

Given:

at moon 

\(g'=\frac g6\)

where g' = acceleration due to gravity at the moon

\(New~Time~ Period= T' =2\pi √ {\frac L{g'}}=2\pi √ {\frac {6L}g}\)

 \(T'=√ {6}T\)

As time period for oscillation increases √6 times, therefore speed becomes √6 times slower.

Concept:

When spring is connected in parallel as shown, the equivalent stiffness is the sum of all individual stiffness of spring.

k_eq = k₁ + k₂

The natural frequency ω_n of a spring-mass system is given by:

\(\omega_n=\sqrt{\frac{k_{eq}}{m}}\;\;\; and\;\;\;\omega_n=2\pi f\)

k_eq = equivalent stiffness and m = mass of body.

Calculation:

Given:

k₁ = 4000 N/m, k₂ = 1600 N/m and m = 1.4 kg

Springs are in parallel correction

∴ k_eq = k₁ + k₂ = 4000 + 1600 = 5600 N/m

\(ω_n = \sqrt {\frac{k_{eq}}{{m}}}\Rightarrow \sqrt {\frac{{5600}}{{1.4}}} = \sqrt {4000} \)

\(f = \frac{ω_n}{{2\pi }} \Rightarrow \frac{{\sqrt {4000} }}{{2\pi }} \approx 10\;Hz\)

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.

Explanation:

Vibration: A body is said to be vibrating if it has a to and fro motion.

Various types of vibrations are mentioned in the table below.

Explanation:-

Critical or whirling speed of a shaft

• When the rotational speed of the system coincides with the natural frequency of lateral/transverse vibrations, the shaft tends to bow out with a large amplitude. This speed is termed as critical/whirling speed.
• Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.
• In other words, the whirling or critical speed is the speed at which resonance occurs.
• Hence we can say that Whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.
• It is the speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite is known as critical or whirling speed.

Deflection of the shaft due to transverse vibration of the shaft.

\(y = \frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1\;}}\)

At critical speed,

\(\omega = {\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{g}{\delta }} \)

where,

ω = Angular velocity of the shaft, k = Stiffness of shaft, e = initial eccentricity of the center of mass of the rotor

m = mass of rotor, y = additional of rotor due to centrifugal force.

Concept:

Transmissibility is given by:

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

\(C = 2\zeta \sqrt {KM} \Rightarrow \zeta = \frac{C}{{2\sqrt {KM} }}\)

Calculation:

After additional damper in parallel \(\zeta ' = \frac{{C + C'}}{{2\sqrt {KM} }}\)

Thus  \(\zeta ' > \zeta \)

Now,  \({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}} \)  

 \(\zeta ' > \zeta \)indicating that \(\omega _d' < {\omega _d}\)

\(time\ period\left( {{T_d}} \right) = \frac{{2\pi }}{{{\omega _d}}}\)

Concept:

When both rotors are free to oscillate the node will be formed at the center (when the rotor is similar) but when one of the rotors is fixed the node will shift to the fixed end.

Calculation:

In the two-rotor system shown in figure

ω₁ = ω₂

\(\sqrt {\frac{{{K_{t1}}}}{{{J_1}}}} = \sqrt {\frac{{{K_{t2}}}}{{{J_2}}}} \)

\({K_t} = \frac{{G{I_p}}}{L}\)

where, \({K_t}\) = torsional stiffness

We have, depending on the position of the node effective length will be

• equal to L/2 when both rotors are free to oscillate.
• equal to L when one of the rotors is fixed.

Hence, by fixing the rotor,

• effective length increases.
• K_t value will decrease.
• natural frequency ω_n will decrease.