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Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B'. The stiffness of spring 'B' will be
  • a)
    one-sixteenth
  • b)
    one- eighth
  • c)
    one-fourth
  • d)
    one-half
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Two closely coiled helical springs 'A' and 'B' are equ...
Explanation:
Stiffness of a spring is defined as the force required to produce unit displacement in the spring. Mathematically,
Stiffness (k) = F/x
where F is the force applied and x is the displacement produced.

Now, let's consider two springs A and B with the following properties:
- Both springs are equal in all respects except for the diameter of wire.
- The diameter of wire of spring A is double that of spring B.

From the above information, we can conclude that:
- The number of coils in both springs will be the same.
- The length of both springs will be the same.
- The material of both springs will be the same.

Since the diameter of wire of spring A is double that of spring B, we can write:
- The radius of wire of spring A is twice that of spring B.
- The cross-sectional area of wire of spring A is four times that of spring B.

Now, let's find the stiffness of both springs using the formula:
k = (Gd^4)/(64D^3n)
where G is the modulus of rigidity of the material, d is the diameter of wire, D is the mean diameter of the spring, and n is the number of coils.

We know that both springs have the same material, number of coils, and length. Therefore, the only variables that differ are the diameter of wire and mean diameter.

Let's assume that the mean diameter of both springs is D. Then, we can write:
- The mean diameter of spring A is D + 2r, where r is the radius of wire.
- The mean diameter of spring B is D + r.

Now, let's find the stiffness of spring A:
kA = (G(2r)^4)/(64(D+2r)^3n)
kA = (G16r^4)/(64(D+2r)^3n)
kA = (G)r^4/(4(D+2r)^3n)

Similarly, let's find the stiffness of spring B:
kB = (G(r)^4)/(64(D+r)^3n)

Now, let's compare the stiffness of both springs:
kA/kB = [(G)r^4/(4(D+2r)^3n)]/[(G(r)^4)/(64(D+r)^3n)]
kA/kB = (1/16)(D+r)^3/(D+2r)^3
kA/kB = (1/16)[(D+r)/(D+2r)]^3

Since the mean diameter of both springs is the same, we can write:
(D+r)/(D+2r) = 1 + (r/D)
Approximating this to first order, we get:
(D+r)/(D+2r) ≈ 1 + (r/D)

Substituting this in the above equation, we get:
kA/kB ≈ (1/16)(1+(r/D))^3

Since r/D is very small (since r is much smaller than D), we can approximate (1+r/D)^3 to 1+3(r/D). Therefore,
kA/kB ≈ (1/16)(1+3(r/D))
kA/kB ≈ (1/16) + (3/16)(r/D)

Since r/D is very small, we can ignore the second term and write:
k
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Two closely coiled helical springs 'A' and 'B' are equ...
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Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B'. The stiffness of spring 'B' will bea)one-sixteenthb)one- eighthc)one-fourthd)one-halfCorrect answer is option 'A'. Can you explain this answer?
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