An object is said to be in the periodic motion when it is repeating its motion with a defined cycle. This type of motion is also referred to as oscillation. Simple examples are the movement of springs and pendulums with many other situations in which oscillations occur. The object has a stable equilibrium position is one important feature of periodic motion. Also, a restoring force is working for string, due to which string may have potential energy. This topic will explain the spring force as well as the spring potential energy formula with examples. Let us learn it!

**Concept of Spring Potential Energy**

Spring is a common tool and usually, their inertia is frequently neglected due to negligible mass. It’s a casual activity that spring when strained will undergo the displacement due to compression. Then it comes to its equilibrium position. Therefore, spring exerts an equal as well as an opposite force on a body that compresses or stretches it.

It is the energy, stored in a compressible or stretchable object like a spring or rubber band or molecule. It’s another name is Elastic potential energy. It is equal to the force times the distance of movement.

If the usual position i.e. without stretched, there is no energy in the spring. But, when we alter its usual position from its usual position, the spring will be able to store energy by the virtue of its position. This energy is termed as potential energy. This is the result of the deformation of a particular elastic object, or a spring. It also describes the work done to stretch the spring. It depends on the spring constant ‘k’ and the distance stretched.

**Potential Energy of a Spring Formula**

String potential energy = force × distance of displacement.

Also, the force is equal to the spring constant × displacement.

The Potential energy of a spring is:

\(P.E = \frac{1}{2} k \times x^{2}\)

P.E. | The potential energy of a spring |

k | Spring constant |

x | Spring displacement |

**Solved Examples for Spring Potential Energy Formula**

Q.1: Find out the potential energy of a spring having spring constant as \(200 Nm^{-1}\)and the displacement is 0.8 m.

Solution: Given parameters are,

Spring constant, \(k = 200N m^{-1},\)

Displacement, x = 0.8m

Potential energy of a string formula is given as:

\(P.E = \frac{1}{2} k \times x^{2}\)

\(P.E = \frac{1}{2} 200 \times (0.8)^{2}\)

= 64 J

Thus, potential energy will be 64 joules.

Q.2: The spring constant of a stretched string is \(50 Nm ^{-1}\) and displacement is 20 cm. Compute potential energy stored in the stretched string.

Solution: Given parameters are,

\(k = 50 Nm ^{-1}\)

x = 20 cm = 0.2 m

Potential energy will be:

\(P.E = \frac{1}{2} k \times x^{2}\)

\(= \frac{1}{2} 50 \times (0.2)^{2}\)

= 1 J

Thus, potential energy stored in the stretched string will be 1 joule.

Q.3: We attach a spring to a board, and use 3 J energy to stretch the spring 99 cm. What be the value of the spring constant using Spring Potential Energy Formula?

Solution: We know that,

Work done = P.E. = 3 J

And displacement, x = 99 cm = 0.99 m.

So, by using formula,

\(P.E = \frac{1}{2} k \times x^ {2}\)

Rearranging it,

\(k = \frac {2 \times P.E.}{x^{2}}\)

\(k = \frac{2 \times 3}{(0.99)^{2}}\)

\(= 6.122 N m^{-1}\)

Thus spring constant will be \(6.122 N m^{-1}.\)

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…