An object is in the periodic motion when its motion repeats with a defined cycle. This type of motion is also oscillation. Simple examples are the movement of springs and pendulums, but there are many other situations in which oscillations occur. An important feature of periodic motion is that the object has a stable equilibrium position. Also, a restoring force which is directed toward that position. Spring has the application of force. This topic will explain the spring force concept and spring force formula with examples. Let us learn it!

**Concept of Spring Force**

For the springs, there is a position at which they are at rest. But when they are either stretched or compressed, then there is a restoring force that will always point in the direction of the equilibrium position. Pendulums are stable during straight down hanging. If a pendulum is pulled away from their equilibrium position, then it swings back and forth as it is acted on by tension force and gravity.

Spring is a tool used usually and their inertia are frequently neglected die to negligible mass. It’s an extremely casual activity that spring, when strained will undergo the displacement when it is compacted it gets compressed. Then it comes to its equilibrium position. Thus, spring exerts an equal as well as an opposite force on a body that compresses or stretches it.

Imagine one end of a spring is attached to a hook, and the other is attached to an object with mass m and allowed to hang down vertically. In this case, the object will have two forces. One force will be the restoring force of the spring, directed upward. The other force will be the force of gravity acting on the mass, directed downward. If the mass is not moving, it will hang at rest at an equilibrium position where the net force is zero,

**The Formula for Spring Force:**

Simple Harmonic Motion comes under periodic motion. In SHM, the restoring force \(F_x\) is directly proportional to the displacement x. This restoring force and the displacement always have opposite signs. A constant of proportionality k makes it possible to form the equation for the force,

i.e. \(F_x\) = – k x

For springs, this relationship is Hooke’s Law, where k will be the spring constant. The SI unit of k is Newton per meter,

In another form, \(F = k (x – x_0)\)

Where,

x | the displacement of the spring from its equilibrium position |

k | the spring constant |

F | spring force |

\(x_o\) | equilibrium position |

**Solved Examples for Spring Force Formula**

Q.1: A spring has a length of 22 cm per s. If it is loaded with 2 kg weight, then it gets stretched by 38 cm per sec. Determine its spring constant using Spring Force Formula.

Solution: Known parameters are,

(Mass) m = 2 kg,

(initial length) \(x_o\) = 22 cm,

(displacement) x = 38 cm

Final displacement = \(x – x_o = 38 cm – 22 cm = 16 cm\) = 0.16 m

The spring force will be,

F = ma (Newton’s law)

= 2 kg × 0.16 m

= 0.32 N

The spring constant,

\(K = – \frac {F}{x- x_0}\)

= – \(\frac {0.32}{0.16}\)

= – 2 N per m

Thus, the spring constant will be – 2 N per m.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…