# Vibration

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 5
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
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Question 5 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 6
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
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Question 6 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}
 Question 7
A hollow spherical ball of radius 20 cm floats in still water, with half of its volume submerged. Taking the density of water as 1000 $kg/m^3$, and the acceleration due to gravity as 10 $m/s^2$, the natural frequency of small oscillations of the ball, normal to the water surface is _________ radians/s (roundoff to 2 decimal places).
 A 2.56 B 4.85 C 8.66 D 9.56
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Question 7 Explanation:
\begin{aligned} \text { Given data: } & R=20 \mathrm{cm}=0.20 \mathrm{m} \\ & \rho=1000 \mathrm{kg} / \mathrm{m}^{3} \\ & g=10 \mathrm{m} / \mathrm{s}^{2} \\ &\left[V=\text { Volume displaced }=\frac{V_{s}}{2}\right] \\ & m g=F_{B}=\rho \psi \cdot g \\ & m=\rho \cdot \frac{V_{s}}{2} \\ V_{s} \rho_{s} & =\rho \cdot \frac{V_{s}}{2} \\ & \rho_{s}=\frac{\rho}{2}=\frac{1000}{2}=500 \mathrm{kg} / \mathrm{m}^{3} \end{aligned}  \begin{aligned} m \ddot{x}+f_{\text {Bextra }}^{\prime} &=0 \\ m \ddot{x}+V_{\text {extra }} \cdot \rho \cdot g &=0 \\ \rho \cdot \frac{V_{s}}{2} \ddot{x}+\left(\pi R^{2} \cdot x\right) \rho \cdot g &=0 \\ \frac{4 \pi R^{3}}{3} \hat{x}+\pi R^{2} x g &=0 \\ \ddot{x}+\frac{3 g}{2 R} x &=0 \\ \omega_{n} &=\sqrt{\frac{3 g}{2 R}}=\sqrt{\frac{3 \times 10}{2 \times 0.20}}=\sqrt{\frac{30}{0.4}} \\ \omega_{n}&=\sqrt{\frac{300}{4}}=\sqrt{75}=8.66 \mathrm{rad} / \mathrm{s} \end{aligned}
 Question 8
The equation of motion of a spring-mass-damper system is given by
$\frac{d^2x}{dt^2}+3\frac{dx}{dt}+9x=10 \sin (5t)$
The damping factor for the system is
 A 0.25 B 0.5 C 2 D 3
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Question 8 Explanation:
$\frac{d^{2} x}{d t^{2}}+3\left(\frac{d x}{d t}\right)+9 x=10 \sin 5 t$
Comparing with standard equation:
$\ddot{x}+\left(2 \xi \omega_{n}\right) \dot{x}+\left(\omega_{n}^{2}\right) x=\left(\frac{F_{o}}{m}\right) \sin \omega t$
\begin{aligned} 2 \xi \omega_{n} &=3 \\ 2 \xi \times 3 &=3 &\left[\begin{array}{l} \omega_{n}^{2}=9 \\ \omega_{n}=3 \end{array}\right] \\ \xi &=\frac{1}{2}=0.5 \end{aligned}
 Question 9
A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut. Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration is
 A $\sqrt{\frac{3}{2}}\sqrt{\frac{k}{m}}$ B $\frac{3}{\sqrt{2}}\sqrt{\frac{k}{m}}$ C $\sqrt{3}\sqrt{\frac{k}{m}}$ D $\sqrt{\frac{k}{m}}$
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Question 9 Explanation: \begin{aligned} k'&= 2k\left ( \frac{l/2}{l} \right )^2\\ k' &=\frac{2k}{4}=\frac{k}{2} \\ k'' &=k\left ( \frac{r}{r} \right )^2=k \\ k_{eq}&=k'+k'' \\ &=\frac{k}{2} +k=\frac{3k}{2}\\ \omega _n &\frac{k_{eq}}{m}=\sqrt{\frac{3k}{2m}} \end{aligned}
 Question 10
A single-degree-of-freedom oscillator is subjected to harmonic excitation $F(t) = F_0 \cos (\omega t)$ as shown in the figure. The non-zero value of $\omega$, for which the amplitude of the force transmitted to the ground will be $F_0$, is
 A $\sqrt{\frac{k}{2m}}$ B $\sqrt{\frac{k}{m}}$ C $\sqrt{\frac{2k}{m}}$ D $2 \sqrt{\frac{k}{m}}$
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Question 10 Explanation:
Given,
$F_{1}=F_{0}$
then transmissibility
$\epsilon =\frac{F_{T}}{F_{0}}=\frac{F_{0}}{F_{0}}=1$
\begin{aligned} \frac{\sqrt{1+\left ( \frac{2\xi \omega }{\omega _n} \right )^2}}{\left ( 1-\left ( \frac{\omega }{\omega _n} \right )^2 \right )^2+\left ( \frac{2\xi \omega }{\omega _n} \right )^2}=1 \\ 1+ \left ( \frac{2\xi \omega }{\omega _n} \right )^2=\left ( 1-\left ( \frac{\omega }{\omega _n} \right )^2 \right )^2+ \left ( \frac{2\xi \omega }{\omega _n} \right )^2\\ 1-\left ( \frac{\omega }{\omega _n} \right )^2 =\pm 1 \\ \text{ Taking (-ve) sign,} \\ 1-\left ( \frac{\omega }{\omega _n} \right )^2 =-1 \\ \frac{\omega }{\omega _n} =\sqrt{2}\\ \omega=\sqrt{2}\omega_n=\sqrt{\frac{2k}{m}} \end{aligned}
There are 10 questions to complete. 