Wire diameter, mean coil diameter and number of turns of a closely-coiled steel spring are d, D and N respectively and stiffness of the spring is k. A second spring is made of the same steel but with wire diameter, mean coil diameter and number of turns as 2d, 2D and 2N respectively. The stiffness of the new spring is
(a) k (b) 2k (c) 4k (d) 8k.

Question

Accepted Answer

(a)

EXPLANATION

We know that stiffness of a close-coiled helical spring is given by

[latex] k=\frac{C d^4}{64 R^3 imes n} [/latex]

1st case, [latex] k_1=\frac{C d^4}{64\left(\frac{D}{2} ight)^3 imes N} \quad\left( ext { here } R=\frac{D}{2} ext { and } n=N ight) [/latex]

2nd case, [latex] k_2=\frac{C(2 d)^4}{64\left(\frac{2 D}{2} ight)^3 imes 2 N} [/latex]

(here dia. of wire = 2d ; Mean coil dia. = 2D and number of turns = 2N)

[latex] =\frac{C d^4 imes 16}{64 imes\left(\frac{D}{2} ight)^3 imes 8 imes 2 N}=\frac{C d^4}{64\left(\frac{D}{2} ight) imes N}= \pmb{k _1} \quad \quad \left\lgroup\because \quad k_1=\frac{C d^4}{\left.64\left(\frac{D}{2} ight)^3 imes N ight)} ight group [/latex]

Question 26.3.46: Wire diameter, mean coil diameter and number of turns of a c...

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