Transmissibility and Magnification Factor MCQ Quiz - Objective Question with Answer for Transmissibility and Magnification Factor - Download Free PDF
Last updated on Apr 3, 2025
Latest Transmissibility and Magnification Factor MCQ Objective Questions
Transmissibility and Magnification Factor Question 1:
A refrigerator unit having a mass of 44 kg is to be supported on 3 springs (of same stiffness = k). If the unit operates at 450 rpm , what will be the stiffness (= k) if only 10% of the shaking force is allowed to be transmitted to the supporting structure? [assume, π2 = 10]
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 1 Detailed Solution
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
Transmissibility and Magnification Factor Question 2:
Consider the following statements. Transmissibility of vibrations
(1) Is more than 1, when ω/ωn < √2
(2) Is less than 1, when ω/ωn > √2
(3) Increases as the damping is increased
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 2 Detailed Solution
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/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.
- If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping
Transmissibility and Magnification Factor Question 3:
"The ratio of the amplitude of the steady-state response to the static deflection is known as Magnification Factor". With respect to the above definition, which of the following statements is incorrect. [Take r as the ratio of forced frequency to natural frequency (ω/ωn)].
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 3 Detailed Solution
Explanation:
The steady-state amplitude for the system:
\(A = \dfrac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
Static deflection (Xst):
\(X_{st}=\dfrac{F}{k}\)
Magnification Factor:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
where \(r=\dfrac{\omega}{\omega_n}\).
Thus M.F = f(r, ξ)
Option 1:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
At r = 1 (resonance)
\(M.F= \dfrac{{1}}{{{2ξ}}}\)
Thus M.F is only a function of the damping ratio i.e. ξ.
Option 2:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
At r = 1 and ξ = 0
\(M.F= \dfrac{{1}}{{{2ξ}}}=\infty\)
The magnification factor is infinite.
Option 3:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
For a given ξ, the value of r at which the M.F is maximum is called ropt.
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
When the denominator is minimum, MF will be maximum.
\(\dfrac{d}{dr}\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}}=0\)
\(r_{opt}=\sqrt{1-2\xi^2}\;\)
Transmissibility and Magnification Factor Question 4:
Consider the following statements. Transmissibility of vibrations
(1) Is more than 1, when ω/ωn < √2
(2) Is less than 1, when ω/ωn > √2
(3) Increases as the damping is increased
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 4 Detailed Solution
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/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.
- If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping
Transmissibility and Magnification Factor Question 5:
"The ratio of the amplitude of the steady-state response to the static deflection is known as Magnification Factor". With respect to the above definition, which of the following statements is incorrect. [Take r as the ratio of forced frequency to natural frequency (ω/ωn)].
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 5 Detailed Solution
Explanation:
The steady-state amplitude for the system:
\(A = \dfrac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
Static deflection (Xst):
\(X_{st}=\dfrac{F}{k}\)
Magnification Factor:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
where \(r=\dfrac{\omega}{\omega_n}\).
Thus M.F = f(r, ξ)
Option 1:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
At r = 1 (resonance)
\(M.F= \dfrac{{1}}{{{2ξ}}}\)
Thus M.F is only a function of the damping ratio i.e. ξ.
Option 2:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
At r = 1 and ξ = 0
\(M.F= \dfrac{{1}}{{{2ξ}}}=\infty\)
The magnification factor is infinite.
Option 3:
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
For a given ξ, the value of r at which the M.F is maximum is called ropt.
\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)
When the denominator is minimum, MF will be maximum.
\(\dfrac{d}{dr}\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}}=0\)
\(r_{opt}=\sqrt{1-2\xi^2}\;\)
Top Transmissibility and Magnification Factor MCQ Objective Questions
For a lightly damped vibrating system executing steady forced vibration, the phase lag of displacement, with respect to force at resonance is
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 6 Detailed Solution
Download Solution PDFExplanation:
At resonance (ω = ωn), phase angle ϕ is 90°.
When ω/ωn ≫ 1; the phase angle is very close to 180°. Here inertia force increases very rapidly and its magnitude is very large.
A single-degree-of freedom oscillator is subjected to harmonic excitation F(t) = F0 cos(ωt) as shown in the figure.
The non-zero value of ω, for which the amplitude of the force transmitted to the ground will be F0, is
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 7 Detailed Solution
Download Solution PDFConcept:
In vibration system, transmissibility is given by
\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)
Where FT = 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)
FT = FO
\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)
∴ 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/ωn)2 = 2
\(\frac{\omega }{{{\omega }_{n}}}=\sqrt{2}\)
∴ ω = √2ωn
∵ Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)
\(\omega =\sqrt{\frac{2K}{m}}\)Consider the following statements. Transmissibility of vibrations
(1) Is more than 1, when ω/ωn < √2
(2) Is less than 1, when ω/ωn > √2
(3) Increases as the damping is increased
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 8 Detailed Solution
Download Solution PDFConcept:
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.
- If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping
A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10-4 seconds. The amplitude in mm of the resulting free vibration is
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 9 Detailed Solution
Download Solution PDFConcept:
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
The ratio of maximum displacement of forced vibration to the deflection due to the static force is known as
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 10 Detailed Solution
Download Solution PDFExplanation:
The ratio of maximum displacement of forced vibration to the deflection due to the static force (F) is known as the Magnification factor.
Amplitude of the steady state response
\(A = \frac{{\frac{{{F_0}}}{k}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }} = \frac{{{X_s}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
Magnification factor:
\(\frac{A}{{{X_s}}} = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)
\(MF\; = \;\frac{1}{{\sqrt {{{\left( {1 - \frac{{{\omega ^2}}}{{\omega _n^2}}} \right)}^2}\; + \;{{\left( {\frac{{2{\rm{\zeta \omega }}}}{{{\omega _n}}}} \right)}^2}} }}\)
The ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio ϵ.
\(\varepsilon = \frac{{\sqrt {1 + {{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2} + {{\left( {1 - \;\frac{{{\omega ^2}}}{{{\omega _n}^2}}} \right)}^2}\;} }}\)
Logarithmic decrement is defined as the natural logarithm of the amplitude reduction factor. The amplitude reduction factor is the ratio of any two successive amplitudes on the same side of the mean position.
\(\delta = {\log _e}\left( {\frac{{{x_n}}}{{{x_{n + 1}}}}} \right)\)
The amplitude of the vibration is the maximum displacement of a vibrating body from the equilibrium position.
In vibration isolation system, if ω/ωn = \(\sqrt{2}\) then transmissibility will be
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 11 Detailed Solution
Download Solution PDFExplanation:
Transmissibility ratio (TR):
In a vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.
\(TR = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)
- 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 = \(\sqrt{2}\), then all the curves pass through the point TR = 1 for all values of damping factor ξ.
- When frequency ratio ω/ωn < \(\sqrt{2}\), 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 > \(\sqrt{2}\), 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 > \(\sqrt{2}\). Here the force transmitted to the foundation increases as the damping is increased.
A machine of 250 kg mass is supported on springs of total stiffness 100 kN/m. Machine has an unbalanced rotating force of 350 N at speed of 3600 rpm. Assuming a damping factor of 0.15, the value of transmissibility ratio is:
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 12 Detailed Solution
Download Solution PDF\(\omega = \frac{{2\pi \times 3600}}{{60}} = 377rad/s\)
Natural frequency, \({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{100 \times 1000}}{{250}}} = 20\ rad/s\)
Now, \(r = \frac{\omega }{{{\omega _n}}} = \frac{{377}}{{20}} = 18.85\)
Transmissibility ratio \(\left( {TR} \right) = \frac{{\sqrt {1 + {{\left( {2\xi r} \right)}^2}} }}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r } \right)}^2}} }}\)
\(= \frac{{\sqrt {1 + {{\left( {2 \times 0.15 \times 18.85} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {18.85} \right)}^2}} \right]^2 + {{\left( {2\times0.15 \times 18.85} \right)}^2}} }}\)
= 0.0162
In a forced vibration system, for which value of frequency ratio (ωf/ωn), the transmissibility is same for all the values of damping factors
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 13 Detailed Solution
Download Solution PDFConcept:
In vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.
\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)
- 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.
A machine of mass m = 200 kg is supported on two mounts, each of stiffness k = 10 kN/m. The machine is subjected to an external force (in N) F(t) = 50 cos 5t. Assuming only vertical translatory motion, the magnitude of the dynamic force (in N) transmitted from each mount to the ground is ______ (correct to two decimal places).
Answer (Detailed Solution Below) 33.00 - 33.50
Transmissibility and Magnification Factor Question 14 Detailed Solution
Download Solution PDFConcept:
Transmissibility is defined as the ratio of force transmitted to the foundation (FT) to the disturbing force (F).
\(TR = \frac{{{F_T}}}{{{F_o}}} = \frac{{\sqrt {1 + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {{{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} - 1} \right]}^2} + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}\)
\({\omega _n} = \sqrt {\frac{k}{m}}\)
For parallel connection of the springs
\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{{k_1} + {k_2}}}{m}} \)
Calculation:
Given, m = 200 kg
F(t) = 50 cos 5t = Fo cos ωt
Fo = 50, ω = 5, ξ = 0
\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{{k_1} + {k_2}}}{m}} = \sqrt {\frac{{2k}}{m}} = \sqrt {\frac{{2 \times 10 \times 1000}}{{200}}} = 10\;rad/s\)
\(r = \frac{\omega }{{{\omega _n}}} = \frac{5}{{10}} = 0.5\)
For ξ = 0
\(TR = \frac{{{F_T}}}{{{F_o}}} = \frac{1}{{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}}} = \frac{1}{{1 - {{\left( {0.5} \right)}^2}}} = 1.333\)
FT = 1.333 × Fo = 1.333 × 50 = 66.67 N
Hence the force transmitted to each mount will be,
FT /2 = 66.67/2 = 33.33 N
In a single degree of freedom underdamped spring mass system as shown in figure, an additional damper is added in parallel such that the system still remains underdamped.
The statement which always remains true is
Answer (Detailed Solution Below)
Transmissibility and Magnification Factor Question 15 Detailed Solution
Download Solution PDFExplanation:
Adding an additional damper in parallel with the existing damper in an underdamped spring-mass system will increase the damping in the system. This increased damping will lead to a decrease in the amplitude of the system's free oscillations and an increase in the time period of free oscillation.
Concept used:
\(ω_d = \sqrt{1 - ξ^2} . ω_n~\)
where
ωd = frequency of damped vibration, ωn = nature frequency of vibration, ξ = damping ratio
we know the Time period of oscillation is given by
\(T_d = \frac{2 \pi}{ω_d}\)
Why adding a damper in parallel increases the time period of free oscillation in a single-degree-of-freedom underdamped spring-mass system:
In a spring-mass system, the time period of free oscillation depends on the system's natural frequency and damping ratio. The natural frequency determines the inherent rate at which the system oscillates without any external influence, while the damping ratio affects how quickly the oscillations decay over time.
Adding a damper in parallel to the existing damper effectively increases the total damping coefficient (c) of the system. While the mass (m) and spring constant (k) remain the same, the increase in c directly impacts the damping ratio (ζ) as follows:
⇒ As the damping coefficient increases, the damping ratio also increases.
⇒ ωd will decrease and as ωd ↓ Td ↑ Since they are inversely proportional.
Important Points
Transmissibility \(ε=\frac{\sqrt{1+\left(\frac{2 ξ \omega}{\omega_{n}}\right)^{2}}}{\sqrt{\left[1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right]^{2}+\left[\frac{2 ξ \omega}{\omega_{n}}\right]^{2}}}\)
In the underdamped system
ε ↓ if ξ ↑ for \(\frac{\omega}{\omega_n} < \sqrt 2\)
ε ↑ if ξ ↑ for \(\frac{\omega}{\omega_n} > \sqrt 2\)
ε will remain the same if \(\frac{\omega}{\omega_n} = \sqrt 2\)
Hence ε depends on ξ as well as \(\frac{\omega}{\omega_n}\) and there is no information provided related to \(\frac{\omega}{\omega_n}\)
so options (1) and (3) will not be our answer.
Hence the correct answer is option (4) Time period of oscillation will increase.