In this article, we will find the Dimensional Formula of Spring constant<\/p>\n\n\n\n
Dimensional formula for is<\/p>\n\n\n\n
$[M^1L^0T^{-2}]$<\/p>\n\n\n\n
Where L<\/strong> -> Length<\/p>\n\n\n\n T<\/strong> -> Time Spring constant is given as per the Hooke’s law as k-> Spring constant of the spring<\/p>\n\n\n\n So,<\/p>\n\n\n\n $k= \\frac {F}{x}$<\/p>\n\n\n\n Now the dimension of displacement= $[L^1]$<\/p>\n\n\n\n Lets derive the dimension of Force<\/p>\n\n\n\n $F= ma$ Where m-> mass<\/p>\n\n\n\n a -> Acceleration<\/p>\n\n\n\n Dimension of Mass = $[M^1]$<\/p>\n\n\n\n Now acceleration $\\text {Dimension of Force} =[M^1] \\times [M^0 L^1T^{-2}] = [M^1L^1T^{-2}]$<\/p>\n\n\n\n Now we know both the displacement and Force dimension , we can calculate the spring constant dimension easily as<\/p>\n\n\n\n $\\text {dimension of spring constant} = \\frac { \\text {dimension of force}} { \\text {dimension of displacement}}$<\/p>\n\n\n\n $= \\frac {[M^1L^1T^{-2}]}{[L^1]} = [M^1L^0T^{-2}]$<\/p>\n\n\n\n Unit of Spring constant is Newton\/meter.<\/p>\n\n\n\n Try the free Quiz given below to check your knowledge of Dimension Analysis:-<\/p>\n\n\n\n
M<\/strong> -> Mass<\/p>\n\n\n\n
We would now derive this dimensional formula.<\/p>\n\n\n\nDerivation for expression of Dimension of Spring Constant<\/h3>\n\n\n\n
$F= kx$
Where x -> displacement of the spring
F -> Force applied on the Spring<\/p>\n\n\n\n
Now<\/p>\n\n\n\n
$a = \\frac {\\Delta v}{t}$
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<\/p>\n\n\n\n
\nQuiz on Dimensional Analysis<\/strong><\/span><\/h4>\n
\n