Simple Mass System MCQ Quiz in தமிழ் - Objective Question with Answer for Simple Mass System - இலவச PDF ஐப் பதிவிறக்கவும்

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,

⇒ 

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,

⇒ 

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, 

So, 

k_s = 4.8 N/mm = 4800 N/m

Then natural frequency is,  

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, 

⇒ 

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

⇒ ≈ 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,  

Natural Linear Frequency (or Natural cycle based Frequency), 

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

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

∴

∴

∴ 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

Comparing the above equation with ẍ + ω_n²x = 0 

∴ 

Explanation:

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

   and  

where k = spring stiffness and m = mass 

If spring stiffness is halved and mass is doubled

then the natural frequency is given by

Hence, the natural frequency will become half.

Concept:

Logarithmic decrement 

Calculation:

ξ = 0.3

∴ δ = 1.976

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

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

Expansion of damper:

Expansion of spring with stiffness k is

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

Comparing with: m_eqθ̈ + c_eqθ̇ +k_eqθ = 0

Also for critical damping ξ = 1

Explanation:

Taking mass moment of inertia about 0

Balancing Torque about 0

Concept:

mass moment of inertia

, 

Then, 

Alternate Method

By Torque Method,

Equ. 1 gives,