Force Constant of Spring : Series and Parallel Combination

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What is the Unit of the Force Constant K of a Spring?

One end of a spring is fixed to a rigid support and a stretching force F is applied at its other end. As a result, the spring elongates by a length x. According to Hooke’s law,
F ∝ x or, F = kx ……. (1)
[where k is a proportionality constant]
According to Newton’s third law of motion, the spring also exerts an equal but opposite reaction F_e.
F_e is called the elastic force of the spring. Therefore, F_e = -F.
So, from equation (1), we get, F_e = -kx ……… (2)
Here, the negative sign indicates that x and F_e are oppositely directed.
The constant k in the above equation is called the force constant of the spring. Now, for x = 1, F = k. So, the force constant of a spring (or, simply, the spring constant) is defined as the external force required to produce a unit elongation.

Units of force constant:

The force constant of a spring measures the stiffness of the spring. The higher the value of k, the stiffer is the spring.
1 N ᐧ m^-1 = \(\frac{10^5 \mathrm{dyn}}{10^2 \mathrm{~cm}}\) = 10³ dyn ᐧ cm^-1

Force Constant of Spring Combination

Series combination: Let the spring constants of two massless springs A and B be k₁ and k₂ respectively [Fig.]. These two springs are joined in series and hung from a rigid support. A downward force F is applied at the free end of spring B. Let the elongations in springs A and B be x₁ and x₂ respectively. Hence, from equation (1), we get,
F = k₁x₁ = k₂x₂
or F = \(\frac{x_1}{\frac{1}{k_1}}\) = \(\frac{x_2}{\frac{1}{k_2}}\) = \(\frac{x_1+x_2}{\frac{1}{k_1}+\frac{1}{k_2}}\) ……….. (3)

Now, the total elongation in this combi-nation of springs is (x₁ + x₂). If k is the equivalent spring constant of this combi-nation, then,
F = k(x₁ + x₂) = \(\frac{x_1+x_2}{\frac{1}{k}}\) ……… (4)
From (3) and (4)
\(\frac{1}{k}\) = \(\frac{1}{k_1}\) + \(\frac{1}{k_2}\) or k = \(\frac{k_1 k_2}{k_1+k_2}\)
The relation shows k < k₁, k₂. This means that series combinations reduce the effective spring constant.

Parallel Combination: Let the spring constants of two massless springs A and B be k₁ and k₂ respectively [Fig.]. These two springs are placed in parallel and hung from a rigid support. Next, a downward force F is applied to the combination so that the springs elongate.

Here, the forces on A and B are F₁ and F₂ respectively (where, F = F₁ + F₂) and the elongation of each is x. So, from equation (1), we get,
F₁ = A₁x and F₂ = k₂x
∴ F₁ + F₂ = (k₁ + k₂)x or, F = (k₁ + k₂)x …….. (5)
Now, the total elongation of this combination of springs is x. If k is the equivalent spring constant of this combination, then,
F = kx …….. (6)
From (5) and (6), we get, k = k₁ + k₂.
This shows that parallel combinations increase the effective spring constant.

Force constant and length of a spring: Let, a spring of length l has a force constant k. If a force F produces an elongation x, then F = kx, or, k = \(\frac{F}{x}\). Now, let us consider half the length, i.e., \(\frac{l}{2}\) of the same spring. As the length is half, the elongation will also be half, i.e., \(\frac{x}{2}\), due to the same force F. The new force constant of this half-spring is k’ = \(\frac{\frac{F}{x}}{\frac{x}{2}}\) = \(\frac{2 F}{x}\) = 2k, i.e., the spring constant is doubled.
Therefore, the spring constant is inversely proportional to its length, i.e., k ∝ \(\frac{1}{l}\), or, kl = constant.

Energy stored in a stretched spring: If a force F = kx stretches a spring from x to x + dx, the work done, dW = Fdx = kxdx. So, the total work done in stretching the spring from elongation 0 to l is W = ∫dW = k\(\int_0^l x d x\) = \(\frac{1}{2} k l^2\). This is stored in the stretched spring as its potential energy. Therefore, the energy stored = \(\frac{1}{2} k l^2\).

Numerical Examples

Example 1.
When a mass of 4 kg is hung from the lower end of a spring, it elongates by 1 cm.
(i) What is the force constant of the spring?
(ii) If a load of 2 kg is hung from the lower end of the spring, then find its elongation.
Solution:
i) We know that, F = kx or, k = \(\frac{F}{x}\)
Here, F = 4 × 9.8 N ; x = 1 cm = 0.01 m
∴ k = \(\frac{4 \times 9.8}{0.01}\) = 3920 N ᐧ m^-1

ii) We know that, F = kx or, x = \(\frac{F}{k}\)
Here, F = 2 × 9.8 N; k = 3920 N ᐧ m^-1
∴ x = \(\frac{2 \times 9.8}{3920}\) = 0.005 m

Example 2.
The force constant of a spring is k. The spring is cut into three equal parts. Find the force constant of each part.
Solution:
Let us consider that the spring elongates by x when a force F is applied on it. So, the Force constant of the spring, k = \(\frac{F}{x}\)
Now, if the spring is cut into three equal parts, then on the application of the same force F, each part of the spring will elongate by \(\frac{x}{3}\).
Therefore, the force constant of each part,
k’ = \(\frac{F}{\frac{F}{3}}\) = \(\frac{3 F}{x}\) = 3k

Example 3.
The force constant of a spring of length l is k. The spring is cut into two parts of lengths l₁ and l₂. If l₁ = nl₂, then find the spring constants k₁, and k₂ of the two parts, n is an integer.
Solution:
According to the problem,
l = l₁ + l₂ = l₁ + \(\frac{l_1}{n}\) = l₁\(\left(\frac{n+1}{n}\right)\)
Again, l = l₁ + l₂ = nl₂ + l₂ = l₂(n + 1)
We know that for a particular spring, the force constant is inversely proportional to the length.
∴ k ∝ \(\frac{1}{l}\) or, kl = constant
∴ k₁l₁ = kl = kl₁\(\frac{(n+1)}{n}\) or, k₁ = \(\frac{k}{n}\)(n + 1)
Similarly, k₂l₂ = kl = kl₂(n + 1) or, k₂ = k(n + 1)