Theory of Machine

Number of links = 4
Number of I-centers 4c_2=6 \omega _2=5 \; rad/sec
\omega _4=?
For
\begin{aligned} V_{I_{24}}&=I_{12}I_{24}\omega _{2}=I_{14}I_{24}\omega_4\\ &=2 \times 5 =8 \times \omega_4\\ \omega_4&=1.25 \; rad/sec \end{aligned}

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

m_1=1kg, m_2=1kg, r_1=1cm, r_2=1cm
Balancing mass, m_b=\sqrt{2}kg
Making force polygon for complete balance

From the right angle triangle,
\begin{aligned} (m_br_b)^2&=(m_1r_1)^2+(m_2r_2)^2 \\ (\sqrt{2}r_b)^2&=(1\times 1)^2+(1\times 1)^2 \\ 2r_b^2&=2 \\ r_b&=1 \end{aligned}

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}

The spin vector will chase the couple on torque vector and produce precision in system.
Hence precision will be -y direction. Rotate slowly (compared to \omega ) about negative z-axis direction.

Speed of gear 1 N_1 =0 ,
Speed of gear 3 N_3 =900rpm ,
Speed of Arm N_{arm} =? ,
Teeth of gear 1 z_1=?
Teeth of gear 2 z_2=20
Teeth of gear 3 z_3=80
Teeth of gear 2
\begin{aligned} z_3&=z_1+2z_2\\ 80&=z_1+2(20)\\ z_1&=40 \end{aligned}
\begin{array}{|c|c|c|c|c|c|} \hline \text{S.No}&\text{Condition of mtotion} &\text{Speed of arm} &\text{Speed of gear 1} &\text{Speed of gear 2} &\text{Speed of gear 3}\\ \hline 1&\text{Arm is fixed and gear 1 with +x rev}&0&+1&\frac{-z_1}{z_2}(1)&\frac{-z_1}{z_3}(1)\\ \hline 2&\text{Arm is fixed and gear 1 with +x rev}&0&+x&\frac{-z_1}{z_2}(x)&\frac{-z_1}{z_3}(x)\\ \hline 3&\text{Arm with +y rev}&+y&y&+y&+y\\ \hline 4&\text{Total }&y&x+y &y-x\frac{z_1}{z_2}&y-\frac{z_1}{z_3}(x)\\ \hline \end{array}
\begin{aligned} N_1&=x+y=0\\ \Rightarrow x&=-y \\ N_3&=y-\frac{z_1}{z_3}(x)=900\\ &=y-\frac{40}{80}(x)=900\\ \Rightarrow y-\frac{40}{80}(-y)=900\\ 1.5y&=900\\ y&=600 rpm \end{aligned}

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

The given mechanism is

As we know sliding pair is a special. Case of turning pair with infinite lengths limit. So the equivalent diagram would be. Since two parallel lines meets at infinite point O, 2O_2 are same.

l_1=3 cm shortest line,
l_2=5cm
l_3=l_{3(\infty )}=L_x+3 , longest link
l_4=l_{4(\infty )}=L_x+1.5
For Grashoff's rule to satisfy
l_1+l_3 \leq l_2 +l_4
3+L_x+3 \leq 5+L_x+1.5
6 \leq 6.5
LHS is less than RHS.
Hence, Grashoff's rule is satisfied in this mechanism. Since shortest link is fixed. It will be a double crank mechanism.

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}

h_1 = 400 mm, h_2 = ?
\omega _1=\omega \;\;\;\omega _2=2\omega
For Watt governor,
\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}