This set of Mechanical Vibrations Multiple Choice Questions & Answers (MCQs) focuses on “Spring Elements”.
1. Given spring constant is 1300 kn/m and deformation produced in the spring is 0.05 m. Find the force produced in the spring
a) 65000 N
b) 65000 KN
c) 75000 N
d) 75000 KN
View Answer
Explanation: The spring force is equal to product of spring constant and deformation in the spring.
F = kx
Given k = 1300 kn/m = (1300 ×103)N/m
x = 0.05 m
F = (1300 × 103)N/m × 0.05 m = 65000
2. The resultant force acting on the following block is
a) Sum of the forces developed in the parallel springs
b) Multiplication of the forces developed in the parallel springs
c) Difference of the forces developed in the first two springs of the parallel springs
d) Difference of the forced developed in the last two springs of the parallel springs
View Answer
Explanation: The springs in above figure are in parallel. The displacement of each spring in the system is same, but the resultant force acting on the block is the sum of the forces developed in the parallel springs.
3. The forces developed in the each spring in following diagram is
a) not equal to the forces acting on the block
b) equal to the forces acting on the block
c) equal to the sum of the forces developed in the series springs
d) equal to the multiplication of the forces developed in the series springs
View Answer
Explanation: The spring in above diagram are in series. The forces developed in each spring is the same and equal to force acting on the block.
4. The displacement of the block in the following diagram is the
a) Multiplication of changes in length of the springs in the series combination
b) Multiplication of changes in spring constants of the springs in the series combination
c) sum of the changes in length of the springs in the series combination
d) sum of the changes in spring constants of the springs in the series combination
View Answer
Explanation: If xi is the change in length of the ith spring, then
x= x1 + x2 + x3 + …….+ xn
5. Given k1 = 20 kn/m, k2 = 30 kn/m , k3 = 40 kn/m are in parallel. The equivalent spring constant is
a) 70 kn/m
b) 80 kn/m
c) 90 kn/m
d) 100 kn/m
View Answer
Explanation: If k1, k2, k3 are in parallel, The equivalent spring constant keq is k1 + k2 + k3 .
Substituting the given values, we have keq = 20 + 30 + 40 = 90 kn/m
6. Given k1 = 90 kn/m, k2 = 30 kn/m , k3 = 60 kn/m are in parallel and static deflection is 0.56 m. The weight of block m is
a) 458
b) 500
c) 1000.8
d) 100.8
View Answer
Explanation: If k1, k2, k3 are in parallel, The equivalent spring constant keq is k1 + k2 + k3 .
Substituting the given values, we have keq= 90 + 30 + 60 = 180 kn/m
Weight of block m is W = δkeq = 0.56 × 180 = 100.8
7. Given k1 = 20 kn/m, k2 = 30 kn/m , k3 = 40 kn/m are in series. The equivalent spring constant is
a) 40 kn/m
b) 20 kn/m
c) 10 kn/m
d) 25 kn/m
View AnswerAnswer: a
Explanation: If given springs are in series, then equivalent spring constant is 1/keq = 1/k1 + 1/k2 + 1/k3 Substituting the values in above equation, we have keq= 40 kn/m.
8. ———— is a link in mechanical system where application of a torque leads to an angular displacement
a) Non linear spring
b) Torsional spring
c) Helical spring
d) Linear spring
View Answer
Explanation: A linear torsional spring has a relationship between an applied moment M and the angular displacement ɵ of M = ktɵ, where kt is torsional spring.
9. The spring force is conservative
a) True
b) False
View Answer
Explanation: Since the work depends upon the initial and final position of the point of application of the spring force and not the path of the system, the spring force is conservative.
10. The strain produced in the rod is——–. ( Displacement = 0.15 m, L= 3m )
a) 0.03
b) 0.02
c) 0.05
d) 0.01
View Answer
Explanation: The application of force to the block of mass m results in displacement x.The strain in rod is F/AE = x/ L = 0.15/3 = 0.05
Sanfoundry Global Education & Learning Series – Mechanical Vibrations.
To practice all areas of Mechanical Vibrations, here is complete set of Multiple Choice Questions and Answers.
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