Transmissibility and Magnification Factor MCQ Quiz in తెలుగు - Objective Question with Answer for Transmissibility and Magnification Factor - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

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:

${\omega }_d=\sqrt{1-{\xi }^2}.{\omega }_n$

where

ω_d = frequency of damped vibration, ωn = nature frequency of vibration, ξ = damping ratio

we know the Time period of oscillation is given by

$T_d=\frac{2\pi }{{\omega }_d}$

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.

⇒ ω_d will decrease and as ω_d ↓ T_d ↑ Since they are inversely proportional.

Important Points

Transmissibility  $\epsilon =\frac{\sqrt{1+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}{\sqrt{{[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2]}^2+{\left[\frac{2\xi \omega }{{\omega }_n}\right]}^2}}$

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.

Concept:

Transmissibility is defined as the ratio of force transmitted to the foundation (FT) to the disturbing force (F).

$T=\frac{F_T}{F_{un}}=\frac{\sqrt{1+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}{\sqrt{{(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

${\omega }_n=\sqrt{\frac{K_{eq}}{m}}$

• 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.

Calculation:

Given:

Forced frequency ω = 40 rad/s, m = 100 kg

We know force transmitted is directly proportional to the transmissibility ratio, more is the transmissibility ratio more is force transmitted.

The transmissibility ratio is proportional to the frequency ratio so we find the frequency ratio of each isolator.

1) For k = 640 kN/m, ζ = 0.70

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{640\times {10}^3}{100}}=80rad/sec$

$\frac{\omega }{{\omega }_n}=\frac{40}{80}=0.5$

2) For k = 640 kN/m, ζ = 0.07

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{640\times {10}^3}{100}}=80rad/sec$

$\frac{\omega }{{\omega }_n}=\frac{40}{80}=0.5$

3) For k = 22.5 kN/m, ζ = 0.70

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{22.5\times {10}^3}{100}}=15rad/sec$

$\frac{\omega }{{\omega }_n}=\frac{40}{15}=2.66$

4) For k = 22.5 kN/m, ζ = 0.070

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{22.5\times {10}^3}{100}}=15rad/sec$

$\frac{\omega }{{\omega }_n}=\frac{40}{15}=2.66$

For the different values of ζ  and frequency ratios, the force transmitted graph has been plotted.

The isolators in the ascending order of the force transmitted to the foundation are 4-3-1-2. 

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

$T=\frac{\sqrt{1+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}{\sqrt{{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2+{\left(1-\frac{{\omega }^2}{{{\omega }_n}^2}\right)}^2}}$

$\frac{\omega }{{\omega }_n}=\sqrt{2}$ or $\frac{\omega }{{\omega }_n}=0$then T = 1 for all values of damping factor c/c_c.

T is infinity at resonance for undamped systems. So at resonance, damping is important.

T is nearly unity at small excitation frequencies

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.

$T.R=\frac{F_t}{F_0}=\frac{\sqrt{1+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }^n}\right)}^2\right]}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

• 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. 

Concept:

Magnification factor:

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

Calculation:

At resonance the ω/ωn= 1

then Magnification Factor becomes, $MF=\frac{1}{2\xi }$

Damping ratio

$\xi =\frac{C}{C_c}$

Critical damping coefficient:

$c_c=2m{\omega }_n=2\sqrt{sm}$

Now, Magnification factor,

$MF=\frac{1}{2\xi }=\frac{1}{\frac{2c}{c_c}}=\frac{1}{\frac{2c}{2\sqrt{sm}}}=\frac{\sqrt{sm}}{c}$

⇒ $MF=\frac{\sqrt{sm}}{c}\times \frac{\sqrt{s}}{\sqrt{s}}=\frac{s}{c\sqrt{\frac{s}{m}}}=\frac{s}{c{\omega }_n}$

Important Formulas:

Amplitude of the steady state response

$A=\frac{\frac{F_0}{s}}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right]}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}=\frac{X_s}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right]}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

Transmissibility:

$T.R=\frac{F_T}{F}=\frac{\sqrt{1+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right]}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

Concept:

$X=\frac{\frac{m_{0}r{\omega }^2}{k}}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}\mathrm{a}\mathrm{n}\mathrm{d}{\omega }_n=\sqrt{\frac{k}{M}}$

For large ω, the ratio ω/ω_n ≫ 1

$\therefore X=\frac{m_{0}r}{M}$

Calculation:

$X=\frac{m_{0}r}{M}$

$X=\frac{1\times 1}{100}$

∴ X = 0.01 cm   

Concept:

Steady state Amplitude:

$A=\frac{F_0/k}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

With ξ = 0

$A=\frac{\frac{F_0}{k}}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_0}\right)}^2\right)}^2}}=\frac{\frac{F_0}{k}}{\left(1-{\left(\frac{\omega }{{\omega }_0}\right)}^2\right)}$

Calculation:

Given:

A = 50 mm, k = 3000 N/m.

F(t) = 100 cos (100t)

ω = 100 rod/s; F₀ = 100 N

${\omega }_n=\sqrt{\frac{k}{m}}$

$A=\frac{\frac{F_0}{k}}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_0}\right)}^2\right)}^2}}=\frac{\frac{F_0}{k}}{\left(1-{\left(\frac{\omega }{{\omega }_0}\right)}^2\right)}$

$0.05=\frac{\frac{100}{3000}}{\left[1-{\left(\frac{100}{{\omega }_0}\right)}^2\right]}$

ω₀ =  173.2 rad/sec

$\Rightarrow {\omega }_0=\sqrt{\frac{k}{m}}=173.2\Rightarrow m=0.1kg$

Explanation:

The steady-state amplitude for the system:

$A={\displaystyle \frac{\frac{F}{k}}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}}$

Static deflection (X_st):

$X_{st}={\displaystyle \frac{F}{k}}$

Magnification Factor:

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}}$

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

where $r={\displaystyle \frac{\omega }{{\omega }_n}}$.

Thus M.F = f(r, ξ)

Option 1:

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

At r = 1 (resonance)

$M.F={\displaystyle \frac{1}{2\xi }}$

Thus M.F is only a function of the damping ratio i.e. ξ.

Option 2:

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

At r = 1 and ξ = 0

$M.F={\displaystyle \frac{1}{2\xi }}=\mathrm{\infty }$

The magnification factor is infinite.

Option 3:

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

For a given ξ, the value of r at which the M.F is maximum is called r_opt.

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

When the denominator is minimum, MF will be maximum.

${\displaystyle \frac{d}{dr}}\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}=0$

$r_{opt}=\sqrt{1-2{\xi }^2}$

Phase angle:

$\varphi ={\tan }^{-1}\left(\frac{2\xi \frac{\omega }{{\omega }_n}}{1-{\left(\frac{\omega }{{\omega }_n}\right)}^2}\right)$

For ω/ωn > 1 the phase will be 180˚ as the value 1- (ω/ω_n)2 will become negative which means out of phase.

• Phase angle varies from zero at low frequencies to 180° at very high frequencies. It changes very rapidly near the resonance and is 90° at resonance irrespective of damping.
• The phase angle between 0 to 90° for all the values of ω/ωn = 0 to 1 and between 90° to 180° for all the values of ω/ωn > 0
• In absence of any damping, phase angle suddenly changes from zero to 180° at resonance.
• The phase angle φ is zero for all the values of damping ratio at ω/ωn = 0

Concept:

$\mathrm{M}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{u}\mathrm{d}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{v}\mathrm{i}\mathrm{b}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}:{\mathrm{x}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{F\left(Disturbingforce\right)}{m\left({\omega }_n^2-{\omega }^2\right)}$

To find ω:

$N=\frac{60\omega }{2\pi }$

$Forcetransmitted\left(F_T\right)=\frac{kF}{k-m{\omega }^2}\dots \left(1\right)$

m = Total engine mass

∵ Only piston will create unbalanced force (F),  

∴ F = m_p ω2 r      ...(2) 

r = radius of the crank = 0.5L, L = Stroke length, ω = Agular frequency of unbalanced damping force, m_p = mass of the piston 

Calculation:

Given:

r = L/2 = 0.075 m, F_T = 700 N, k = 344000 N/m, m_p = 3.5 kg.

Using equation (2),

F = 0.2625 ω²

Using equation (1),

$\Rightarrow 700=\frac{344000\times 0.2625{\omega }^2}{344000-200{\omega }^2}$

$\therefore 2.408\times {10}^8-140000{\omega }^2=90300{\omega }^2$

∴ ω = 32.33 rad/s

$N=\frac{60w}{2\pi }$

∴ N = 308.78 rpm