Question

Two helical tensile spring of the same material and also having identical mean coil diameter and weight, have wire diameters d and \(\frac d2\). The ratio of their stiffness is
1. 1
2. 4
3. 64
4. 128

Answer 1

Correct Answer - Option 3 : 64

Concept:

Spring constant / Stiffness of a spring (k): 

Force required to produce unit deflection.

\(k = \frac{P}{\delta }\)

where P = Load & δ = Deflection.

Deflection of spring in terms of active turns:

\(\delta = \frac{{8P{D^3}N}}{{G{d^4}}}\)

\(∴ k = \frac{P}{{\frac{{8P{D^3}N}}{{G{d^4}}}}} ⇒ \frac{{G{d^4}}}{{8{D^3}N}}\)

where D = Diameter of the spring coil, d = Diameter of spring wire, N = No. of active turns, and G = Modulus of Rigidity.

Calculation:

Given:

G₁ = G₂, D₁ = D₂, d₁ = d, d₂ = \(\frac d2\), W₁ = W₂

\(∴ k = \frac{{G{d^4}}}{{8{D^3}N}}\)

\(\frac{k_1}{k_2}=\left(\frac{d_1}{d_2}\right)^4\times\left(\frac{N_2}{N_1}\right)\)

Weight (W) = ρVg

i.e. W ∝ V     [∵ ρ₁ = ρ₂]

Volume = Area × Length ⇒ \(\frac{\pi}{4}d^2\;\times\;(\pi DN)\)

W₁ = W₂ ⇒ V₁ = V₂

\(d_1^2N_1=d_2^2N_2\)

\(\frac{N_2}{N_1}=\left(\frac{d_1}{d_2}\right)^2\Rightarrow 4\)

\(\frac{k_1}{k_2}=\left(\frac{d_1}{d_2}\right)^4\times\left(\frac{N_2}{N_1}\right)\)

\(\frac{k_1}{k_2}=\left(2\right)^4\times4\Rightarrow 64\)