# Vibration

 Question 1
The figure shows a block of mass m = 20 kg attached to a pair of identical linear springs, each having a spring constant k = 1000 N/m. The block oscillates on a frictionless horizontal surface. Assuming free vibrations, the time taken by the block to complete ten oscillations is _________ seconds. (Rounded off to two decimal places)
Take $\pi= 3.14$ A 2.36 B 5.35 C 8.25 D 6.28
GATE ME 2023   Theory of Machine
Question 1 Explanation:
$K_{eq}=K+K=2K=2 \times 1000=2000 \mathrm{~N} / \mathrm{m}$
$\mathrm{m}=20 \mathrm{~kg}$
So, $\quad \omega_{n}=\sqrt{\frac{K_{\text {eq }}}{m}}=\sqrt{\frac{2000}{20}}=10 \mathrm{rad} / \mathrm{sec}$
So, $\quad T_{n}=\frac{2 \pi}{\omega_{n}}=\frac{2 \pi}{10}$
Time taken for 10 oscillations $=10 \times T_{n}=10 \times \frac{2 \pi}{\omega_{n}}=10 \times \frac{2 \pi}{10}=6.28 \mathrm{sec}$
 Question 2
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 2 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 3
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
GATE ME 2022 SET-2   Theory of Machine
Question 3 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 4
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 4 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 5
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
GATE ME 2022 SET-1   Theory of Machine
Question 5 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} \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}