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Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.

## 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} = \(\frac{10^5 \mathrm{dyn}}{10^2 \mathrm{~cm}}\) = 10^{3} dyn ᐧ cm^{-1}

### Force Constant of Spring Combination

Series combination: Let the spring constants of two massless springs A and B be k_{1} and k_{2} 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 x_{1} and x_{2} respectively. Hence, from equation (1), we get,

F = k_{1}x_{1} = k_{2}x_{2}

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 (x_{1} + x_{2}). If k is the equivalent spring constant of this combi-nation, then,

F = k(x_{1} + x_{2}) = \(\frac{x_1+x_2}{\frac{1}{k}}\) ……… (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 < k_{1}, k_{2}. This means that series combinations reduce the effective spring constant.

Parallel Combination: Let the spring constants of two massless springs A and B be k_{1} and k_{2} 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 F_{1} and F_{2} respectively (where, F = F_{1} + F_{2}) and the elongation of each is x. So, from equation (1), we get,

F_{1} = A_{1}x and F_{2} = k_{2}x

∴ F_{1} + F_{2} = (k_{1} + k_{2})x or, F = (k_{1} + k_{2})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 = k_{1} + k_{2}.

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 l_{1} and l_{2}. If l_{1} = nl_{2}, then find the spring constants k_{1}, and k_{2} of the two parts, n is an integer.

**Solution:**

According to the problem,

l = l_{1} + l_{2} = l_{1} + \(\frac{l_1}{n}\) = l_{1}\(\left(\frac{n+1}{n}\right)\)

Again, l = l_{1} + l_{2} = nl_{2} + l_{2} = l_{2}(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

∴ k_{1}l_{1} = kl = kl_{1}\(\frac{(n+1)}{n}\) or, k_{1} = \(\frac{k}{n}\)(n + 1)

Similarly, k_{2}l_{2} = kl = kl_{2}(n + 1) or, k_{2} = k(n + 1)