Vibration

Question 1
A spring mass damper system (mass m, stiffness k, and damping coefficient c) excited by a force F(t)=B\sin \omega t , where B,\omega, t are the amplitude, frequency and time, respectively, is shown in the figure. Four different responses of the system (marked as (i) to (iv)) are shown just to the right of the system figure. In the figures of the responses, A is the amplitude of response shown in red color and the dashed lines indicate its envelope. The responses represent only the qualitative trend and those are not drawn to any specific scale.

Four different parameter and forcing conditions are mentioned below.

(P)\;\;C \gt 0 \text{ and }\omega =\sqrt{k/m}
(Q)\;\;C \lt 0 \text{ and }\omega \neq 0
(R)\;\;C=0 \text{ and }\omega =\sqrt{k/m}
(S)\;\;C=0 \text{ and }\omega \cong \sqrt{k/m}

Which one of the following options gives correct match (indicated by arrow \rightarrow ) of the parameter and forcing conditions to the responses?
A
(P) \rightarrow (i), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (ii)
B
(P) \rightarrow (ii), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (i)
C
(P) \rightarrow (i), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (iii)
D
(P) \rightarrow (iii), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (i)
GATE ME 2022 SET-2   Theory of Machine
Question 1 Explanation: 
FBD of mass

m\ddot{x}+c\dot{x}+kx=F(t)
Solution of differential equation
x(t) = (C.F) + (P.I)
Considering condition (P)
If C > 0 and \omega =\sqrt{k/m}
For this condition the displacement (x) is given by
[Assume the system to be under damped]
x(t)=X_0e^{-\zeta \omega _nt} \cos (\omega _dt -\phi )+ X \cos (\omega t-\phi )
Transient response + steady state response
As t\rightarrow \infty the transient response decays to zero and only steady state response will remain
x(t)=X \cos (\omega t-\phi )
For this condition the response curve will be


Considering condition (Q)
c \gt 0 \text{ and } \omega \neq 0
The differential equation becomes
m\ddot{x}-c\dot{x}+kx=F(t)
Solution of above differential equation is
x(t)=X_0e^{+c_1t} \cos (\omega _dt -\phi )+ X \cos (\omega t-\phi )
As t\rightarrow \infty the transient response approaches to \infty and increases exponentially
The plot will be


Considering condition (R)
C=0,\omega =\sqrt{\frac{k}{m}} \;\;(Resonance)
The differential equation is
m\ddot{x}+kx=F(t)
Solution for above differential equation
x(t)=x_o \cos \omega _nt+\frac{\dot{x_0}}{\omega _n} \sin \omega _n t+\underbrace{\frac{x_{static}\omega _nt}{2} \sin (\omega _n t)}_{\text{Incraeses with time linearly}}
So the correct plot will be


Considering condition (S)
C=0,\omega \cong \sqrt{\frac{k}{m}}
If the force frequency is close to, but not exactly equal to, natural frequency of the system, a phenomenon is known as beating. In this kind of vibration the amplitude buils up and then diminishes in a regular pattern.
The displacement can be expressed as
x(t)=\left ( \frac{B}{m} \right )\left [ \frac{\cos \omega t- \cos \omega _n t}{\omega _n^2-\omega } \right ]
The correct plot will be

Question 2
For a dynamical system governed by the equation,
\ddot{x}(t)+2\varsigma \omega _n\dot{x}(t)+\omega _n^2x(t)=0
the damping ratio \varsigma is equal to \frac{1}{2\pi} \log_e 2 . The displacement x of this system is measured during a hammer test. A displacement peak in the positive displacement direction is measured to be 4 mm. Neglecting higher powers ( > 1) of the damping ratio, the displacement at the next peak in the positive direction will be _______ mm (in integer)
A
1
B
2
C
3
D
4
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Question 2 Explanation: 
For a dynamic system,
\ddot{x}+2\zeta \omega _n\dot{x}+\omega _n^2x=0
The above equation represents a damped free vibration system.
Given damping factor, \zeta=\frac{1}{2 \pi} \ln (2)
Given a peak x_{n-1}=4 \; mm
x_n=?
Logarithmic decrement is given by
\ln\left ( \frac{x_{n-1}}{x_n} \right )=\frac{2 \pi\zeta }{\sqrt{1-\zeta ^2}}
As question is asking to neglect the higher power of \zeta , so neglecting \zeta ^2 , as \zeta \lt 1 .
\begin{aligned} \ln \left ( \frac{x_{n-1}}{x_n} \right )&=2 \pi\zeta \\ 2 \pi\zeta &= \ln (2) \\ \Rightarrow \ln \left ( \frac{x_{n-1}}{x_n} \right ) &= \ln (2)\\ \Rightarrow \frac{4}{x_n}&=2 \\ \Rightarrow x_n&=2\; mm \end{aligned}
Question 3
Consider a forced single degree-of-freedom system governed by
\ddot{x}(t)+2\zeta \omega _n\dot{x}(t)+\omega _n^2x(t)=\omega _n^2 \cos (\omega t) ,
where \zeta and \omega _n are the damping ratio and undamped natural frequency of the system, respectively, while \omega is the forcing frequency. The amplitude of the forced steady state response of this system is given by [(1-r^2)^2+(2\varsigma r)^2]^{-\frac{1}{2}} , where r=\omega /\omega _n . The peak amplitude of this response occurs at a frequency \omega =\omega _p . If \omega _d denotes the damped natural frequency of this system, which one of the following options is true?
A
\omega _p \lt \omega _d \lt \omega _n
B
\omega _p = \omega _d \lt \omega _n
C
\omega _d \lt \omega _n = \omega _p
D
\omega _d \lt \omega _n \lt \omega _p
GATE ME 2022 SET-1   Theory of Machine
Question 3 Explanation: 
The relation between the frequency at peak and natural frequency is given
\omega _p=\omega _n\sqrt{1-2\zeta ^2}
\omega _p \gt \omega _n
The relation between the damped frequency and natural frequency
\omega _d=\omega _n\sqrt{1-\zeta ^2}
\omega _d \gt \omega _n
\frac{\omega _p}{\omega _d}=\frac{\sqrt{1-2\zeta ^2}}{\sqrt{1-\zeta ^2}} \lt 1
Therefore, \omega _p \lt \omega _d \lt \omega _n
Question 4
A rigid uniform annular disc is pivoted on a knife edge A in a uniform gravitational field as shown, such that it can execute small amplitude simple harmonic motion in the plane of the figure without slip at the pivot point. The inner radius r and outer radius R are such that r^2=R^2/2, and the acceleration due to gravity is g. If the time period of small amplitude simple harmonic motion is given by T=\beta \pi \sqrt{R/g}, where \pi is the ratio of circumference to diameter of a circle, then \beta=________ (round off to 2 decimal places).

A
5.25
B
2.65
C
3.75
D
9.86
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Question 4 Explanation: 


Considering the differential ring mass of differential ring
\begin{aligned} d_m&=\frac{\text{mass of ring}}{\text{volume of ring}}(dV)\\ &=\frac{M}{\pi(R^2-r^2)t}2 \pi ada \times t\\ &=\frac{2Mada}{(R^2-r^2)}\\ I_0&=\bar{I}=\int a^2 d_m\\ &=\int_{r}^{R}\frac{2Ma^3da}{(R^2-r^2)}\\ &=\frac{2M}{(R^2-r^2)} \times \left [ \frac{a^4}{4} \right ]_r^R\\ &=\frac{M}{2}\frac{R^4-r^4}{R^2-r^2}\\ &=\frac{M}{2}{R^2+r^2}\\ I&=\bar{I}+mr^2\\ &=\frac{M}{2}(R^2+r^2)+Mr^2=\frac{5MR^2}{4} \end{aligned}


Taking moments about A
\begin{aligned} \Sigma M_A&=0\\ I\ddot{\theta }+Mg(r \sin \theta )&=0\\ \frac{5MR^2}{4}\ddot{\theta }+Mg\left ( \frac{R}{\sqrt{2}} \right )\theta &=0\;\;(\because \sin \theta \cong \theta )\\ \ddot{\theta }+\frac{4g}{5\sqrt{2}R}\theta &=0\\ \omega _n&=\sqrt{\frac{4g}{5\sqrt{2}R}}\\ T&=\frac{2 \pi}{\omega _n}= \pi\left ( 2 \times \sqrt{\frac{5\sqrt{2}}{5} \times \frac{R}{g}} \right )\\ \beta &=2.65 \end{aligned}
Question 5
Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.


The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural frequency of the system is
A
\sqrt{\frac{kr^2}{J-mr^2}}
B
\sqrt{\frac{kr^2}{J+mr^2}}
C
\sqrt{\frac{k}{m}}
D
\sqrt{\frac{kr^2}{J}}
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Question 5 Explanation: 


\begin{aligned} E &=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} k x^{2}+\frac{1}{2} I \omega^{2} \\ &=\frac{1}{2} m r^{2} \dot{\theta}^{2}+\frac{1}{2} k r^{2} \theta^{2}+\frac{1}{2} J \dot{\theta}^{2} \\ &=\frac{1}{2}\left[\left(J+m r^{2}\right) \dot{\theta}^{2}+k r^{2} \theta^{2}\right]=0 \\ \frac{d E}{d t} &=0 \\ \left(J+m r^{2}\right) \times 2 \dot{\theta} \ddot{\theta}+k r^{2} 2 \theta \dot{\theta} &=0 \\ \left(\ddot{\theta}+\frac{k r^{2}}{J+m r^{2}}\right) \theta &=0 \\ \omega_{n} &=\sqrt{\frac{k r^{2}}{J+m r^{2}}} \end{aligned}
Question 6
A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The isolator can be considered as a combination of stiffness (K) and damper (damping factor, \xi) in parallel. The designer has the following four isolators:

1) K = 640 kN/m, \xi = 0.70
2) K = 640 kN/m, \xi = 0.07
3) K = 22.5 kN/m, \xi = 0.70
4) K = 22.5 kN/m, \xi = 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.
A
1-3-4-2
B
1-3-2-4
C
4-3-1-2
D
3-1-2-4
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Question 6 Explanation: 
Transmitted force problem:
\begin{aligned} \omega&=40 \mathrm{rad} / \mathrm{s} \text { (force frequency) }\\ m&=100 \mathrm{~kg} \\ \epsilon&=\frac{\sqrt{1+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}}}{\sqrt{\left\{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}\right\}}} \\ \omega_{n}&=\sqrt{\frac{640 \times 10^{3}}{100}}=80 \Rightarrow \frac{\omega}{\omega_{n}}=0.5\\ \omega_{n}&=\sqrt{\frac{22.5 \times 10^{3}}{100}}=47.434164 \Rightarrow \frac{\omega}{\omega_{n}}=\frac{40}{15}=2.666 \end{aligned}



Question 7
A tappet valve mechanism in an IC engine comprises a rocker arm ABC that is hinged at B as shown in the figure. The rocker is assumed rigid and it oscillates about the hinge B. The mass moment of inertia of the rocker about B is 10^{-4}\;kg.m^2. The rocker arm dimensions are a = 3.5 cm and b = 2.5 cm. A pushrod pushes the rocker at location A, when moved vertically by a cam that rotates at N rpm. The pushrod is assumed massless and has a stiffness of 15 N/mm. At the other end C, the rocker pushes a valve against a spring of stiffness 10 N/mm. The valve is assumed massless and rigid.

Resonance in the rocker system occurs when the cam shaft runs at a speed of ______rpm (round off to the nearest integer).
A
496
B
4739
C
790
D
2369
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Question 7 Explanation: 




I=10^{-4} \mathrm{~kg}-\mathrm{m}^{2}
By D'Alembert Principle
I \ddot{\theta}+\left[10000 \times(0.025)^{2}+15000 \times(0.035)^{2}\right] \theta=0
\begin{aligned} \left(10^{-4}\right) \ddot{\theta}+(24.625) \theta&=0 \\ \ddot{\theta}+\left(\frac{24.625}{10^{-4}}\right) \theta&=0\\ \Rightarrow \qquad\qquad \omega_{n}^{2}&=(246250) \\ \omega_{n}&= 496.2358\; \text{rad/s} \\ \Rightarrow \qquad\qquad N_C&=\frac{496.2358 \times 60}{2 \pi} \\ &=4738.70 \text{ rpm}\end{aligned}

Question 8
Consider a two degree of freedom system as shown in the figure, where PQ is a rigid uniform rod of length, b and mass, m.

Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is
A
\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta
B
\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta
C
\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2}
D
\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta +Fb\sin\theta
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Question 8 Explanation: 


\begin{aligned} \text { For mass, } T&=\frac{1}{2} M \dot{x}^{2}\\ V&=\frac{1}{2} K x^{2} \end{aligned}
For mass, m


\begin{aligned} d m &=\frac{m}{b} d y \\ \text { Displacement } &=x+y \sin \theta \\ \text { Velocity } &=\dot{x}+y \cos \theta \dot{\theta} \\ d T & =\frac{1}{2} d M V e l^{2}=\frac{1}{2} d m\left[\dot{x}^{2}+y^{2} \dot{\theta}^{2} \cos ^{2} \theta+2 \dot{x} \dot{\theta} y \cos \theta\right] \\ T &=d T=\frac{1}{2} \frac{m}{b} \int\left(\dot{x}^{2}+y^{2} \dot{\theta}^{2} \times 1+2 \dot{x} \dot{\theta} y \cos \theta\right) d y \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+2 \dot{x} \dot{\theta} \cos \theta \frac{b^{2}}{2}\right) \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+\dot{x} \dot{\theta} b^{2} \cos \theta\right) \\ T &=\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} \dot{x} \dot{\theta} b \cos \theta \\ V &=-m g \frac{b}{2} \cos \theta \end{aligned} For both masses, M and m
\begin{aligned} T &=\frac{1}{2} M \dot{x}^{2}+\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} b \dot{\theta} \dot{x}^{2} \cos \theta \\ V &=\frac{1}{2} k x^{2}-m g \frac{b}{2} \cos \theta \\ L &=T-V \\ &=\frac{1}{2}(M+m) \dot{x}^{2}+\frac{1}{2} m b \dot{x} \dot{c} \cos \theta+\frac{m b^{2} \dot{\theta}^{2}}{6}-\frac{1}{2} k x^{2}+m g \frac{b}{2} \cos \theta \end{aligned}
Question 9
Consider a single degree of freedom system comprising a mass M, supported on a spring and a dashpot as shown in the figure.

If the amplitude of the free vibration response reduces from 8 mm to 1.5 mm in 3 cycles, the damping ratio of the system is ______ (round off to three decimal places).
A
0.044
B
0.088
C
0.064
D
0.032
GATE ME 2021 SET-1   Theory of Machine
Question 9 Explanation: 


\begin{aligned} \frac{x_{0}}{x_{3}} &=\frac{8}{1.5}=\frac{80}{15} \\ \frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}} &=\frac{80}{15} \\ e^{8} \cdot e^{8} \cdot e^{8} &=\frac{80}{15} \\ e^{38} &=\ln \left(\frac{80}{15}\right)=1.67397 \\ \delta &=0.55799 \\ \frac{2 \pi \xi}{\sqrt{1-\xi^{2}}} &=0.55799 \\ \xi &=0.088 \end{aligned}
Question 10
A rigid block of mass m_1 = 10 kg having velocity v_0 = 2 m/s strikes a stationary block of mass m_2 = 30 kg after travelling 1 m along a frictionless horizontal surface as shown in the figure.

The two masses stick together and jointly move by a distance of 0.25 m further along the same frictionless surface, before they touch the mass-less buffer that is connected to the rigid vertical wall by means of a linear spring having a spring constant k = 10^5 N/m. The maximum deflection of the spring is _________ cm (round off to 2 decimal places).
A
10
B
1
C
100
D
2
GATE ME 2020 SET-2   Theory of Machine
Question 10 Explanation: 
Collision Theory
Conservation of momentum,
\begin{aligned} m_{1} \times v_{0}+m_{2} \times 0 &=\left(m_{1}+m_{2}\right) \times v \\ 10 \times 2 &=(10+30) v \\ 20 &=40 v \\ v &=0.5 \mathrm{m} / \mathrm{s} \\ \text{Now,}\quad \frac{1}{2}\left(m_{1}+m_{2}\right) v^{2} &=\frac{1}{2} k x^{2}\\ \frac{1}{2} \times 40 \times(0.5)^{2} &=\frac{1}{2} \times 10^{5} \times x^{2} \\ 10 &=10^{5} \times x^{2} \\ \Rightarrow \quad x^{2} &=\frac{1}{10^{4}}\\ x&=\frac{1}{100} \mathrm{m}=1 \mathrm{cm} \end{aligned}
There are 10 questions to complete.

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