Question

The close-coiled helical springs 'A' and 'B' are of same material, same coil diameter, same wire diameter and subjected to same load. If the number of turns of spring 'A' is half that of spring 'B', the ratio of deflection of spring 'A' to spring 'B' is

• A. \(\frac{1}{2}\)
• B. 1
• C. 2
• D. 4

Hint:

Deflection ($\delta$) and stiffness ($k$) behaviors in helical springs: - Deflection is directly proportional to the number of turns: \(\delta \propto n\) - Axial spring stiffness is inversely proportional to the number of active turns: \(k = \frac{W}{\delta} = \frac{G d^4}{8 D^3 n} \implies k \propto \frac{1}{n}\) Cutting a spring in half reduces the turns by half, which doubles its stiffness and halves its deflection under the same load.

Answer 1

The Correct Option is A

For a close coiled helical spring, the deflection per single active coil is the same for any spring made of the same wire, same coil diameter, same material and carrying the same load, since none of those quantities differ between spring A and spring B here. So the total deflection of each spring is just this fixed "deflection per turn" multiplied by how many turns it has, in other words deflection is directly proportional to the number of turns. Spring A has only half as many turns as spring B, so its total deflection must also be half of spring B's deflection. The ratio \(\delta_A : \delta_B\) therefore works out to \(1:2\), which is \(\dfrac{1}{2}\), option (1).