Question

A particle in an unidirectional potential field where the potential energy (U) of a particle depends on the x – coordinate given by ${{U}_{x}}=k\left( 1-\cos ax \right)$ and ‘k’ and ‘a’ are constants. Find the physical dimensions of ‘a’ and ‘k’.

Answer 1

Hint: We are given the expression for potential energy of a particle in a unidirectional potential field. We need to find the dimensions of the constants in the expression. There is a cosine function in the expression and we know that the trigonometric expressions do not have dimensions. Thus we can find the dimensions of the two constants.
Formula used:
$U=Fs$

Complete answer:
In the question we are given a particle which is in a unidirectional potential field.
It is said that the potential energy only depends on the x – coordinate and it is given as,
${{U}_{x}}=k\left( 1-\cos ax \right)$
We are asked to find the physical dimensions of the constants ‘a’ and ‘k’ in the above equation.
We know that the trigonometric ratios are dimensionless.
Hence the cosine function in the equation is dimensionless, i.e. the term ‘$ax$’ is dimensionless.
We know that the dimension of ‘x’ will be [L], hence for ‘$ax$’ to be dimensionless the dimension of a should be, $\left[ \dfrac{1}{L} \right]$.
$\begin{align}
  & \left[ ax \right]=\left[ \dfrac{1}{L} \right]\left[ L \right] \\
 & \Rightarrow \left[ ax \right]=\text{dimensionless} \\
\end{align}$
Therefore the dimension of a is $\left[ {{L}^{-1}} \right]$
Now we need to find the dimension of the constant ‘k’.
Since $\left( 1-\cos ax \right)$ is dimensionless, the dimension of ‘k’ will be the dimension of potential energy.
We know that potential energy is given by the equation,
$U=Fs$, where ‘F’ is force and ‘s’ is displacement.
Force, ‘F’ is given as, $F=ma$, where ‘m’ is mass and ‘a’ is acceleration.
Therefore we can write the potential energy as,
$\Rightarrow U=ma\times s$
We know that the dimensions of ‘m’, ‘a’ and ‘s’ is,
$\left[ m \right]=\left[ M \right]$
$\left[ a \right]=\left[ \dfrac{L}{{{T}^{2}}} \right]$
$\left[ s \right]=\left[ L \right]$
Therefore the dimension,
$\Rightarrow \left[ U \right]=\left[ k \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
Therefore the dimension of ‘k' and ‘a’ are,
$\left[ k \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
$\left[ a \right]=\left[ {{L}^{-1}} \right]$

Note:
Dimensions are basically related to units, even though they are independent of the system of units. Dimensional analysis is a method used to relate the dimensions of physical quantities in an equation.
According to the principle of homogeneity, the dimension of every term on the right hand side of the dimensional equation will be the same as the dimension of every term on the left hand side of the equation.