Simple Mass System MCQ Quiz in বাংলা - Objective Question with Answer for Simple Mass System - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:-

Springs in series - 

Consider two springs with force constants k₁ and k₂ connected in series supporting a load,

F = mg. 

The equivalent stiffness of the system can be written as,

⇒ \(\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}\;\)

Springs in Parallel -

Consider two springs with force constants k₁ and k₂ connected in parallel, supporting a load F = mg. 

The equivalent stiffness of the system can be written as,

⇒ k_p = k₁ + k₂ 

Natural frequency for an undamped spring-mass system can be written as,

⇒ \(\omega_n = \sqrt{\frac{k}{m}}\;\)

Calculation:-

Given:-

m = 5 kg, k1 = 8 N/mm, k2 = 12 N/mm

Case(1) - The mass is suspended at the bottom of two springs in series -

Equivalent stiffness is, \(\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}\;\)

So, \(\frac{1}{k_s}=\frac{1}{8}+\frac{1}{12}\;\)

k_s = 4.8 N/mm = 4800 N/m

Then natural frequency is, \(\omega_n = \sqrt{\frac{k_s}{m}}\;\) 

\(\omega_{n1} = \sqrt{\frac{4.8×10^3}{5}}= 30.983\;rad/s\;\)

Case(2) - The mass is fixed between two springs.

Equivalent stiffness is, k_p = k₁ + k₂

So, k_p = 8 + 12 = 20 N/mm = 20 × 10³ N/m

Then natural frequency is, \(\omega_n = \sqrt{\frac{k_p}{m}}\;\)

⇒ \(\omega_{n2}= \sqrt{\frac{20000}{5}}= 63.245\;rad/s\;\)

The ratio of natural frequencies of case (ii) to those of case (i) is,

⇒ \(\frac{\omega _{n2}}{\omega _{n1}}= \frac{63.245}{30.983}=2.04\;\)≈ 2

Concept:

Natural frequency (Eigen frequency): It is the frequency at which a system tends to oscillate indefinitely in the absence of any driving or damping force.

Natural Angular Frequency (or Natural Circular Frequency)is given by, \({\rm{\omega }} = {\rm{\;}}\sqrt {\frac{{\rm{k}}}{{\rm{m}}}} {\rm{\;and}},\) 

Natural Linear Frequency (or Natural cycle based Frequency), \({\rm{\;f}} = \frac{{\rm{\omega }}}{{2{\rm{\pi }}}} = \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{\rm{k}}}{{\rm{m}}}} \)

Where k is the stiffness of the system and m is the mass of the system.

Note:

Natural circular frequency is expressed in radian/sec unit whereas Natural linear frequency is expressed in cycles/sec or Hz unit.

Concept: Consider a spring-mass system

Natural frequency for a spring mass system is

\({\omega _n} = \sqrt {\frac{k}{m}} \)

Calculation:

Given that, k = 2 N/cm = 200 N/m, m = 8 kg

Concept:

By Newton’s law

Iθ̈ + mx²θ̈ + kx2θ = 0

where; I = moment of inertia of pulley, m = mass of box = 5 kg, M = mass of pulley = 20 kg, k = stiffness of spring, r = radius of pulley

Mr²/2 θ̈ + mr²θ̈ + kr²θ  =  0

(10 × r²) θ̈   + (5 × r²) θ̈ + (1500 × r²)θ = 0  

(15 × r²)θ̈  + (1500 × r²) θ = 0

Comparing the above Equation with

θ̈  + ω_n²θ = 0

∴ \(\omega _n^2 = \frac{{1500{r^2}}}{{15{r^2}}}\)

∴ \({\omega _n} = \sqrt {100} \)

∴ ω_n = 10 rad/s

Using Newton's method and taking moment about point O.

m(2a)²ẍ  + k(a)²x = 0

m4a²ẍ + ka²x = 0

\(\mathop {\ddot x}{+} \frac{{k{a^2}}}{{4m{a^2}}}x = 0\)

\(\mathop {\ddot x}{+} \frac{{k{}}}{{4m{}}}x = 0\)

Comparing the above equation with ẍ + ω_n²x = 0 

∴ \({\omega _n} = \sqrt {\frac{k}{{4m}}} \)

Explanation:

The natural frequency of a spring-mass system is given by,

\(f_n = \frac {\omega_n}{2\pi}\)   and  \(\omega_n = \sqrt {\frac km} \)

where k = spring stiffness and m = mass 

If spring stiffness is halved and mass is doubled

then the natural frequency is given by

\(\omega_n' = \sqrt {\frac {0.5k}{2m}}\)

\(\omega_n'= \sqrt {\frac {k}{4m}}\)

\(\omega_n'= \frac 12 \sqrt {\frac {k}{m}}\)

\(\omega_n'=\frac {w_n}{2} \)

\(f_n' = \frac {f_n}{2}\)

Hence, the natural frequency will become half.

Concept:

Logarithmic decrement \(\left( \delta \right) = \frac{{2\pi \xi }}{{\sqrt {1 - {\xi ^2}} }}\)

Calculation:

ξ = 0.3

\(\therefore \delta = \frac{{2\pi\; \times\; 0.3}}{{\sqrt {1\; - \;{{0.3}^2}} }}\)

∴ δ = 1.976

The ratio of 2 successive amplitude is always constant and given as,

\(\frac{{{X_n}}}{{{X_{n + 1}}}} = {e^\delta } = {e^{1.976}}\)

\(\therefore \frac{{{X_n}}}{{{X_{n + 1}}}} = 7.21\)

For small angular rotation θ of the rod, compression in the spring (3k) is

\({\delta _1} = \left( {\frac{L}{4}} \right)\theta \Rightarrow {F_1} = {F_{3k}} = {k_1}{\delta _1} = 3k\frac{L}{4}\theta = \frac{3}{4}kL\theta \)

Expansion of damper:

\({\delta _2} = \left( {\frac{L}{4}} \right)\theta \)

\( \Rightarrow {\dot \delta _2} = \frac{L}{4}\dot \theta \)

\( \Rightarrow {F_2} = {F_c} = C\;{\dot \delta _2} = \frac{{CL}}{4}\dot \theta \)

Expansion of spring with stiffness k is

\({\delta _3} = \left( {\frac{L}{2} + \frac{L}{4}} \right)\theta = \frac{{3L}}{4}\theta \)

\( \Rightarrow {F_3} = {F_k} = \frac{{3L}}{4}\theta k\)

Now taking moment of all forces about O and inertia forces to be zero, we get

\(I\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{3}{4}kL\theta \left( \frac{L}{4} \right)+\frac{CL}{4}\dot{\theta }\frac{L}{4}+\frac{3L}{4}\theta k\left( \frac{3L}{4} \right)=0\)

\(I = {I_{about\;centre}} + m{\left( {\frac{L}{4}} \right)^2} = \frac{{m{L^2}}}{{12}} + \frac{{m{L^2}}}{{16}} = \frac{{7m{L^2}}}{{48}}\)

\(\frac{7}{48}m{{L}^{2}}\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{3}{16}k{{L}^{2}}\theta +\frac{9}{16}k{{L}^{2}}\theta +\frac{C{{L}^{2}}}{16}\dot{\theta }=0\)

\(\Rightarrow \frac{7}{3}m{{L}^{2}}\overset{\ddot{\ }}{\mathop{\theta }}\,+12k{{L}^{2}}\theta +C{{L}^{2}}\dot{\theta }=0\)

Comparing with: m_eqθ̈ + c_eqθ̇ +k_eqθ = 0

\(\Rightarrow \xi =\frac{{{C}_{eq}}}{2\sqrt{{{k}_{eq}}{{m}_{eq}}}}=\frac{C{{L}^{2}}}{2\sqrt{12k{{L}^{2}}\times \frac{7}{3}m{{L}^{2}}}}=\frac{C{{L}^{2}}}{2\sqrt{12k{{L}^{2}}\times \frac{7}{3}m{{L}^{2}}}}=\frac{1}{2\left( 2\sqrt{7} \right)}\frac{C}{\sqrt{mk}}\)

Also for critical damping ξ = 1

\(\Rightarrow \frac{C}{\sqrt{mk}}\times \frac{1}{4\sqrt{7}}=1\)

\(\Rightarrow \frac{C}{\sqrt{mk}}=4\sqrt{7}\)

Explanation:

Taking mass moment of inertia about 0

\(I = \frac{{m{L^2}}}{{12}} + M{\left( {\frac{L}{2} - \frac{L}{3}} \right)^2} + m \times {\left( {\frac{{2L}}{3}} \right)^2}\)

\( = \frac{{M{L^2}}}{{12}} + \frac{{M{L^2}}}{{36}} + \frac{{4M{L^2}}}{9} = \frac{{2M{L^2}}}{9}\)

Balancing Torque about 0

\(I\alpha = K \times \frac{{2L}}{3} \times \left( {\frac{{2L}}{3}\theta } \right) + K \times \frac{L}{3} \times \left( {\frac{L}{3}\theta } \right)\)

\(\frac{{2M{L^2}}}{9}\frac{{{d^2}\theta }}{{d{t^2}}} = \frac{{5k}}{{2M}} = \omega _n^2\theta \)

Concept:

\(I\mathop {\theta }\limits^{..} + k \times \frac{{2l}}{3} \times \frac{{2l}}{3} \times \theta + k \times \frac{l}{3} \times \frac{l}{3} \times \theta = 0\)

\(I\mathop {\theta }\limits^{..} + \left( {\frac{{4k{L^2}}}{9} + \frac{{k{l^2}}}{9}} \right)\theta = 0\)

mass moment of inertia

, \(\begin{array}{l} {I_O} = {I_C} + m{\left( {\frac{{2L}}{3} - \frac{L}{2}} \right)^2}\\ {I_C} = \frac{{m{L^2}}}{{12}}\\ {I_O} = \frac{{m{L^2}}}{{12}} + \frac{{m{L^2}}}{{36}}\\ \therefore I = \frac{{m{L^2}}}{9} \end{array}\)

\(\therefore \frac{{m{l^2}}}{9}\mathop \theta \limits^{..} + \frac{{5k{l^2}}}{9}\theta = 0\)

Then, \({\omega _n} = \sqrt {\frac{{{k_{eq}}}}{{{m_{eq}}}}} \)

\({\omega _n} = \sqrt {\frac{{\frac{{5k{l^2}}}{9}}}{{\frac{{m{l^2}}}{9}}}} = \sqrt {\frac{{5k}}{m}} \)

Alternate Method

By Torque Method,

\(I = \frac{{m{l^2}}}{{12}} + m\;{\left( {\frac{l}{2} - \frac{l}{3}} \right)^2} = \frac{{m{l^2}}}{{12}} + \frac{{m{l^2}}}{{36}} = \frac{{m{l^2}}}{9}\)

\({k_1} = k{\left( {\frac{{2l}}{3}} \right)^2} = \frac{{4{l^2}k}}{9},\;{k_2} = k{\left( {\frac{l}{3}} \right)^2} = \frac{{k{l^2}\;}}{9}\)

\(\therefore k = {k_1} + {k_2} = \frac{{4{l^2}k\;}}{9} + \frac{{k{l^2}}}{9}\)

\(k = \frac{{5k{l^2}}}{9}\;\)

Equ. 1 gives,

\(\left( {\frac{{m{l^2}}}{9}} \right)\mathop {\ddot \theta }\limits + \left( {\frac{{5k{l^2}}}{9}} \right)\theta = 0\)

\(\mathop {{\rm{\ddot \theta }}}\limits + {\rm{}}\left( {\frac{{\frac{{5{\rm{k}}{{\rm{l}}^2}}}{9}}}{{\frac{{{\rm{m}}{{\rm{l}}^2}}}{9}}}} \right){\rm{\theta }} = 0\)

\(\mathop {{\rm{\ddot \theta }}}\limits + {\rm{}}\left( {\frac{{5k}}{m}} \right){\rm{\theta }} = 0\)

\(\therefore {\omega _n} = \sqrt {\frac{k}{m}} = \;\sqrt {\frac{{\frac{{5k}}{m}}}{1}} = \sqrt {\frac{{5k}}{m}} \;\;\)