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

Take \pi= 3.14

2.36 | |

5.35 | |

8.25 | |

6.28 |

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}

\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?

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 |

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 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}

\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?

\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 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

\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).

5.25 | |

2.65 | |

3.75 | |

9.86 |

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}

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}

There are 5 questions to complete.