Question

A helical spring has N turns of coil diameter D and a second spring, made of same wire diameter and same material, has N/2 turns of coil of diameter 2D. If the stiffness of the first spring is k, then the stiffness of the second spring will be

1. k/4
2. k/2
3. 2k
4. 4k

Answer 1

Correct Answer - Option 1 : k/4

Concept:

The stiffness of Spring is given as

\(k = \frac{W}{\delta } = \frac{{G{d^4}}}{{8W{D^3}n}}\)

where d= diameter of spring wire, D = Mean diameter of the spring or coil diameter, G = Shear modulus, n = number of coils, W = Load

If all the parameters remain constant except n, and D

Then,

 \(k\propto \frac 1{D^3 n}\)

Calculation:

Given:

n1 = N and n2 = N/2, k1 = k, D1 = D and D2 = 2D,

\(k\propto \frac 1{D^3 n}\)

\(\frac {k_2}{k_1}=\frac {D_1^3n_1}{D_2^3n_2}\)

\(\frac {k_2}{k_1}=\frac {D^3N}{(2D)^3 N/2}\)

\(\frac {k_2}{k_1}=\frac {D^3N}{8D^3 N/2}\)

\(\frac {k_2}{k_1}=\frac{1}{4}\)

\(k_2=\frac{k_1}{4}=\frac{k}{4}\)