**Dimensional Formula of Spring Constant**

**with its Derivation**

In this article, we will find the dimension of Spring Constant

Dimensional formula for is

$[M^1L^0T^{-2}]$

Where

**M** -> Mass

**L** -> Length

**T** -> Time

We would now derive this dimensional formula.

### Derivation for expression of Dimension of Spring Constant

Spring constant is given as per the Hooke’s law as

$F= kx$

Where x -> displacement of the spring

F -> Force applied on the Spring

k-> Spring constant of the spring

So,

$k= \frac {F}{x}$

Now the dimension of displacement= $[L^1]$

Lets derive the dimension of Force

$F= ma$

Now

Where m-> mass

a -> Acceleration

Dimension of Mass = $[M^1]$

Now acceleration

$a = \frac {\Delta v}{t}$

Now dimension of Velocity= $[M^0 L^1T^{-1}]$

dimension of Time = $[M^0 T^1]$

So dimension of Acceleration = $ \frac {[M^0 L^1T^{-1}]}{ [M^0 T^1]}= [M^0 L^1T^{-2}]$

So, Dimension of force is given by

$\text {Dimension of Force} =[M^1] \times [M^0 L^1T^{-2}] = [M^1L^1T^{-2}]$

Now we know both the displacement and Force dimension , we can calculate the spring constant dimension easily as

$\text {dimension of spring constant} = \frac { \text {dimension of force}} { \text {dimension of displacement}}$

$= \frac {[M^1L^1T^{-2}]}{[L^1]} = [M^1L^0T^{-2}]$

Unit of Spring constant is Newton/meter.

Try the free Quiz given below to check your knowledge of Dimension Analysis:-

**Quiz on Dimensional Analysis**

**Related Articles**

- Dimensional Analysis:- a very good website for physics concepts
- Newton’s Law of Motion
- Newton’s Second Law of Motion

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