A simple spring-mass vibrating system has a natural frequency of $${{\text{f}}_{\text{n}}}.$$ If the spring stiffness is halved and the mass is doubled, then the natural frequency will become

Two closely-coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be __________ that of spring 'B'.

A

One-sixteenth

B

One-eighth

C

One-fourth

D

One-half

Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B' The stiffness of spring 'B' will be __________ that of spring 'A'

A

One-sixteenth

B

One-eighth

C

One-fourth

D

One-half

When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the deflection of the spring ($$\delta $$) is given by (where d = Diameter of spring wire, n = No. of turns of the spring and C = Modulus of rigidity for the spring material)

A

$$\frac{{{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

B

$$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

C

$$\frac{{4{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

D

$$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be

A

$$\frac{\tau }{2}$$

B

$$\tau $$

C

$$2\tau $$

D

$$4\tau $$

The capacity of a spring to store energy is called the spring form co-efficient. Stiffness of a spring is measured by the

A

Ability to absorb shock

B

Capacity to store energy

C

Ratio of the wire & coil diameters

D

Load to produce unit deflection

Vibrating screens are used for handling large tonnages of materials. The vibrating motion is imparted to the screening surface by means of

A

Electromagnets

B

Cams or eccentric shafts

C

Unblanced fly wheels

D

Either A, B or C

Two closely-coiled helical springs 'A' and 'B' of the same material, same number of turns and made from same wire are subjected to an axial load W. The mean diameter of spring 'A' is double the mean diameter of spring 'B'. The ratio of deflections in spring 'B' to spring 'A' will be

A

$$\frac{1}{8}$$

B

$$\frac{1}{4}$$

C

2

D

4

Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B'. The ratio of deflections in spring 'A' to spring 'B' is

A

$$\frac{1}{8}$$

B

$$\frac{1}{4}$$

C

$$\frac{1}{2}$$

D

2

When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)

A

$$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

B

$$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

C

$$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

D

$$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

When a helical compression spring is cut into halves, the stiffness of the resulting spring will be

A

Same

B

Double

C

One-half

D

One-fourth

A simple spring-mass vibrating system has a natural frequency of $${{\text{f}}_{\text{n}}}.$$ If the spring stiffness is halved and the mass is doubled, then the natural frequency will become

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