Question 1 |

In the configuration of the planar four-bar
mechanism at a certain instant as shown in the
figure, the angular velocity of the 2 cm long link is
\omega _2 =5rad/s. Given the dimensions as shown, the
magnitude of the angular velocity \omega _4 of the 4 cm
long link is given by _____ rad/s (round off to 2
decimal places).

1.25 | |

0.75 | |

2.25 | |

3.75 |

Question 1 Explanation:

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}

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?

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?

(P) \rightarrow (i), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (ii) | |

(P) \rightarrow (ii), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (i) | |

(P) \rightarrow (i), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (iii) | |

(P) \rightarrow (iii), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (i) |

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

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 |

A rigid body in the X-Y plane consists of two point
masses (1 kg each) attached to the ends of two
massless rods, each of 1 cm length, as shown in the
figure. It rotates at 30 RPM counter-clockwise about
the Z-axis passing through point O. A point mass of
\sqrt{2} kg, attached to one end of a third massless rod,
is used for balancing the body by attaching the free
end of the rod to point O. The length of the third rod
is ________ cm.

1 | |

\sqrt{2} | |

\frac{1}{\sqrt{2}} | |

\frac{1}{2\sqrt{2}} |

Question 3 Explanation:

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}

Question 4 |

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)

\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)

1 | |

2 | |

3 | |

4 |

Question 4 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}

\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 5 |

A massive uniform rigid circular disc is mounted
on a frictionless bearing at the end E of a massive
uniform rigid shaft AE which is suspended
horizontally in a uniform gravitational field by two
identical light inextensible strings AB and CD as
shown, where G is the center of mass of the shaft-disc assembly and g is the acceleration due to
gravity. The disc is then given a rapid spin \omega about
its axis in the positive x-axis direction as shown,
while the shaft remains at rest. The direction of
rotation is defined by using the right-hand thumb
rule. If the string AB is suddenly cut, assuming
negligible energy dissipation, the shaft AE will

rotate slowly (compared to \omega ) about the negative z-axis direction | |

rotate slowly (compared to \omega ) about the positive z-axis direction | |

rotate slowly (compared to \omega ) about the negative y-axis direction | |

rotate slowly (compared to \omega ) about the positive y-axis direction |

Question 5 Explanation:

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.

Question 6 |

A schematic of an epicyclic gear train is shown in
the figure. The sun (gear 1) and planet (gear 2) are
external, and the ring gear (gear 3) is internal. Gear
1, gear 3 and arm OP are pivoted to the ground at O.
Gear 2 is carried on the arm OP via the pivot joint
at P, and is in mesh with the other two gears. Gear 2
has 20 teeth and gear 3 has 80 teeth. If gear 1 is kept
fixed at 0 rpm and gear 3 rotates at 900 rpm counter
clockwise (ccw), the magnitude of angular velocity
of arm OP is __________rpm (in integer).

300 | |

600 | |

900 | |

1200 |

Question 6 Explanation:

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}

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}

Question 7 |

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?

\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?

\omega _p \lt \omega _d \lt \omega _n | |

\omega _p = \omega _d \lt \omega _n | |

\omega _d \lt \omega _n = \omega _p | |

\omega _d \lt \omega _n \lt \omega _p |

Question 7 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

\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 8 |

A planar four-bar linkage mechanism with 3
revolute kinematic pairs and 1 prismatic kinematic
pair is shown in the figure, where AB\perp CE and FD\perp CE. The T-shaped link CDEF is constructed such
that the slider B can cross the point D, and CE is
sufficiently long. For the given lengths as shown,
the mechanism is

a Grashof chain with links AG, AB, and CDEF completely rotatable about the ground link FG | |

a non-Grashof chain with all oscillating links | |

a Grashof chain with AB completely rotatable about the ground link FG, and oscillatory links AG and CDEF | |

on the border of Grashof and non-Grashof chains with uncertain configuration(s) |

Question 8 Explanation:

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.

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.

Question 9 |

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).

5.25 | |

2.65 | |

3.75 | |

9.86 |

Question 9 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 10 |

The figure shows a schematic of a simple Watt
governor mechanism with the spindle O_1O_2 rotating
at an angular velocity \omega about a vertical axis. The
balls at P and S have equal mass. Assume that there
is no friction anywhere and all other components are
massless and rigid. The vertical distance between
the horizontal plane of rotation of the balls and the
pivot O_1
is denoted by h . The value of h=400 mm
at a certain \omega . If \omega is doubled, the value of h will be
_________ mm.

50 | |

100 | |

150 | |

200 |

Question 10 Explanation:

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}

\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}

There are 10 questions to complete.

figures of questions are missing,and in some question statement is given wrong..

Thank you prakash for your suggestions.

Can you please share the questions numbers for the above suggestions.

question 8

We have review the question 8 with original published paper by IIT. There is only one figure which is there in our question -8. It might possible due to slow internet image takes time to load.

Thank you for your support.

Please check figures of Q28 &29. They are wrong.

what is wrong with q28 and q29?can you say that?

Dear Franklin,

q28 and q29 are correct.

Answer to Q 184 is wrong, it should be option B- 3.

PLEASE ADD ALL GATE 2022 PROBLEMS

YOUR EFFORT IS COUNTLESS