Question

Two helical tensile springs of the same material and also having identical mean coil diameter and weight have wire diameters 'd' and 'd/2'. The ratio of their stiffness is:

(a) 1:2
(b) 1:4
(c) 2:1
(d) 4:1

Answer 1

The stiffness ratio of two springs with wire diameters d and d/2 is 16:1, indicating that the spring with the larger wire diameter is significantly stiffer. There may be a typo in the question as the exact answer is not listed in the provided options.

Step-by-step explanation:

The question concerns the comparison of the stiffness of two helical tensile springs made with the same material and having the same mean coil diameter and weight but different wire diameters. The ratio of wire diameters is 1 to 0.5 (or d to d/2).

Stiffness in springs, often represented by the spring constant (k), depends on the material and the geometry of the spring. According to the formula for stiffness of a helical spring, k is inversely proportional to the fourth power of the wire diameter. The formula is k = (Gd^4) / (8D^3n), where G is the modulus of rigidity, D is the mean diameter of the spring, n is the number of active coils, and d is the diameter of the wire.

For two springs with diameters d and d/2, we compare them as follows:

1. Original spring: k1 = (Gd^4) / (8D^3n)
2. Spring with half diameter: k2 = (G(d/2)^4) / (8D^3n) = (Gd^4) / (128D^3n)
3. So the ratio k2/k1 = (Gd^4) / (128D^3n) divided by (Gd^4) / (8D^3n) equals 1/16

Therefore, the ratio of their stiffness is 1:16. However, this is not one of the options provided, so possibly there is a typo in the question. If the right options were included, the correct answer would be (k1/k2) = 16:1, indicating that the spring with the larger wire diameter is 16 times stiffer than the one with the smaller diameter.

But if we are to choose from the options given, the closest answer (though not accurate) would be 4:1, with the understanding that there's a mistake in the question.

Therefore answer is (d) 4:1.