Simple harmonic motion i.e. SHM is a very interesting type of motion. It is frequently applied in the oscillatory motion of the objects. Springs are usually having SHM. Springs have their own natural “spring constants” which define how stiff they are. Hook’s law is a famous law that explains the SHM and gives a formula for the force applied using spring constant. In this article, we will discuss the spring constant formula with examples. Let us learn the interesting concept!

**Spring Constant Formula**

**What is Hook’s Law? or What is Spring Constant?**

Hook’s law is establishing the relationship between the force applied and the distance stretched in the spring. It says that the force required to compress or extend a spring is directly proportional to the distance it is stretched.

Letter K is spring constant, and it has the units as N/m. According to Newton’s Third Law of Motion, when spring is pulled, it pulls back with a restoring force. This restoring force follows the Hooke’s Law, which relates the force of the spring to the spring constant

**force of the spring = -\( (spring constant k) \ times (displacement) \)**

**F = -k × x**

F | restoring force of the spring (directed toward equilibrium) |

K | spring constant (units N/m) |

x | displacement of the spring from its equilibrium position |

The negative sign shows the reverse direction of the force of reaction.

**Solved Examples**

Q.1: Find the spring constant for spring if it requires a 9000 Newton force to pull spring 30.0 cm from the position of equilibrium.

Solution:

To solve for the spring constant, k, we can rearrange the formula for spring constant as:

F= -K × x

i.e. K = \( \frac{-F}{x} \)

In this example, a 9000 N force is pulling on a spring. It means that the spring pulls back with an equal and opposite force of -9000 N.

Also, the displacement is 30.0 cm = 0.30 m. Thus putting the values in the above formula, we get,

K = \( \frac{-9000}{0.30}\)

i.e. K= 30000 N/m

The spring constant of this spring is 30000 N/m.

Q. 2: A 3500 Newton force is applied to a spring that has a spring constant of k = 14000 N/m. Calculate that how far from equilibrium will the spring be displaced?

Solution: We can find the displacement by rearranging the spring constant formula:

F= -K × x

i.e. x = \( \frac{-F}{K} \)

In this example, a 3500 N force is pulling on a spring. It means that the spring pulls back with an equal and opposite force of -3500 N.

Thus,

x = \( \frac{-3500}{14000}\)

x=0.250 m

x = 25.0 cm

Therefore, the spring is displaced by 25.0 cm.

Q. 3: A spring with load 5 Kg is stretched by 40 cm. Find out its spring constant.**
**Solution: As given in the problem,

Mass of spring, m = 5 Kg

Displacement, x = 40 cm

We know that,

Force F = m × a

Putting the values in the formula,

F = 5 × 0.4

F = 2 Newton

Now, the spring constant is:

K= \( \frac {-F}{x}\)

i.e K = \( \frac{– 2}{0.4} \)

K = – 5 N/m

Therefore the spring constant will be -5 N/m

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