Simple Mass System MCQ Quiz in తెలుగు - Objective Question with Answer for Simple Mass System - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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, ${\omega }_n=\sqrt{\frac{k_s}{m}}$ 

${\omega }_{n1}=\sqrt{\frac{4.8\times {10}^3}{5}}=30.983rad/s$

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, ${\omega }_n=\sqrt{\frac{k_p}{m}}$

⇒ ${\omega }_{n2}=\sqrt{\frac{20000}{5}}=63.245rad/s$

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, $\omega =\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{a}\mathrm{n}\mathrm{d},$ 

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

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

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

∴ ${\omega }_n=\sqrt{100}$

∴ 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

$\ddot{x}+\frac{k{a}^2}{4m{a}^2}x=0$

$\ddot{x}+\frac{k}{4m}x=0$

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  ${\omega }_n=\sqrt{\frac{k}{m}}$

where k = spring stiffness and m = mass 

If spring stiffness is halved and mass is doubled

then the natural frequency is given by

${\omega }_n^{\prime }=\sqrt{\frac{0.5k}{2m}}$

${\omega }_n^{\prime }=\sqrt{\frac{k}{4m}}$

${\omega }_n^{\prime }=\frac{1}{2}\sqrt{\frac{k}{m}}$

${\omega }_n^{\prime }=\frac{w_n}{2}$

$f_n^{\prime }=\frac{f_n}{2}$

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

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

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

$\frac{7}{48}m{L}^2\ddot{\theta }+\frac{3}{16}k{L}^2\theta +\frac{9}{16}k{L}^2\theta +\frac{C{L}^2}{16}\dot{\theta }=0$

$\Rightarrow \frac{7}{3}m{L}^2\ddot{\theta }+12k{L}^2\theta +C{L}^2\dot{\theta }=0$

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:

$I\stackrel{..}{\theta }+k\times \frac{2l}{3}\times \frac{2l}{3}\times \theta +k\times \frac{l}{3}\times \frac{l}{3}\times \theta =0$

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

mass moment of inertia

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

$\therefore \frac{m{l}^2}{9}\stackrel{..}{\theta }+\frac{5k{l}^2}{9}\theta =0$

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,

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

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

$\therefore k=k_1+k_2=\frac{4l^{2}k}{9}+\frac{k{l}^2}{9}$

$k=\frac{5k{l}^2}{9}$

Equ. 1 gives,

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

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

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

$\therefore {\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{\frac{5k}{m}}{1}}=\sqrt{\frac{5k}{m}}$