Question

The potential energy of a particle varies the distance x from a fixed origin as $U=\dfrac{A\sqrt{x}}{{{x}^{2}}+B}$, where A and B are dimensional constants, then find the dimensional formula of AB.

Answer 1

To solve this problem, we must first find the dimension of potential energy given by; $\left[ U \right]=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$and the dimension of (x), given by; $\left[ x \right]=\left[ {{L}^{1}} \right]$. Using the variation given in the problem, we will be able to find the dimensional values of A and B.

Step by step solution:
Let’s start by finding out the default dimension of potential energy. We know that the potential energy of a body of mass (m) kept at a height (h) above the ground and the acceleration due to gravity is (g). Hence, the potential energy (U) is given by; $U=mgh$. The dimensional analysis of the potential energy will be: $U=mgh\Rightarrow \left[ U \right]=\left[ {{M}^{1}} \right]\left[ {{L}^{1}}{{T}^{-2}} \right]\left[ {{L}^{1}} \right]\Rightarrow \left[ U \right]=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$.
Similarly, the dimension of (x) is given by; $\left[ x \right]=\left[ {{L}^{1}} \right]$.
Now, let’s consider the problem given. We have the potential energy given by $U=\dfrac{A\sqrt{x}}{{{x}^{2}}+B}$.
Now using the principle of homogeneity, the dimension of the denominator must be constant. Hence, the dimension of ${{x}^{2}}=B$. Therefore, ${{x}^{2}}=B\Rightarrow \left[ B \right]=\left[ {{x}^{2}} \right]=\left[ {{L}^{2}} \right]$.
Further, the previous value of the potential energy using the dimensional analysis now becomes; $\begin{aligned} & U=\dfrac{A\sqrt{x}}{{{x}^{2}}+B}\Rightarrow \left[ U \right]=\dfrac{\left[ A \right]\left[ \sqrt{x} \right]}{\left[ {{x}^{2}} \right]+\left[ B \right]}\Rightarrow \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\dfrac{\left[ A \right]\left[ {{L}^{\dfrac{1}{2}}} \right]}{\left[ {{L}^{2}} \right]+\left[ {{L}^{2}} \right]} \\\ & \Rightarrow \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\left[ A \right]\left[ {{L}^{\dfrac{1}{2}-2}} \right]\Rightarrow \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\left[ A \right]\left[ {{L}^{\dfrac{-3}{2}}} \right] \\\ \end{aligned}$
Therefore, the dimension of A becomes; $\left[ A \right]=\left[ {{M}^{1}}{{L}^{\dfrac{7}{2}}}{{T}^{-2}} \right]$.
Using the dimensions of A and B as found above, we find the dimension of the product AB to be: $AB=\left[ A \right]\left[ B \right]=\left[ {{M}^{1}}{{L}^{\dfrac{7}{2}}}{{T}^{-2}} \right]\left[ {{L}^{2}} \right]\Rightarrow \left[ AB \right]=\left[ {{M}^{1}}{{L}^{\dfrac{11}{2}}}{{T}^{-2}} \right]$.

Note: The process of dimensional analysis is used extensively to check if a given formula is correct by matching the dimensions of the left hand side and the right hand side dimensionally.
The principle of homogeneity is extremely necessary to solve problems, since for any process of addition or subtraction to occur, the dimensions of both the terms must match for the operation to occur. Hence, the terms in either addition or subtraction operation must be homogeneous with respect to the dimensions.