Question

In the relation $y=r\sin \left( \omega t-kx \right)$, the dimensions of $\dfrac{\omega }{k}$ are
A.$\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$
B.$\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$
C.$\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}} \right]$
D.$\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]$

Answer 1

Hint: You could firstly identify the given expression and thereby identify the given quantities. Then you could recall the expression for these quantities separately and hence find their individual dimensions from it. Now you could use dimensional analysis on the given ratio to find the required dimension.

Formula used: Expression for angular frequency,
$\omega =\dfrac{2\pi }{T}$
Expression for wave number,
$k=\dfrac{2\pi }{\lambda }$

Complete step by step answer:
In the question we are given a relation containing $\omega $ and $k$, and we are asked to find the dimension of $\dfrac{\omega }{k}$.
We know that, the wave displacement formula or the formula representing the displacement of a harmonic wave traveling in positive x-direction is given by,
$y\left( x,t \right)=A\sin \left( kx-\omega t \right)$
Comparing this with the given relation in the question we see that both are the same. Therefore the relation $y=r\sin \left( \omega t-kx \right)$ is a representation of wave- displacement.
‘$r$’ in this expression is the amplitude of the wave that the expression represents, $\omega $ is the angular frequency of the wave and $k$ is the angular wave number.
Let us give a brief description about $\omega $ and $k$.
We know that angular frequency is the angular displacement of any element of the wave per unit time. Also, it is given by the expression,
$\omega =\dfrac{2\pi }{T}$
And also, it has an SI unit of radian per second $\left( rad/s \right)$. From the expression of angular frequency we get its dimension as,
$\left[ \omega \right]={{\left[ T \right]}^{-1}}$ …………………………………… (1)
We know that the angular wave number or simply the wave number of a wave can be defined as the number of radians per unit distance. It is given by the expression,
$k=\dfrac{2\pi }{\lambda }$
The SI unit of angular wave number is radian per meter$\left( rad/m \right)$. From its expression we could write its dimension as,
$\left[ k \right]={{\left[ L \right]}^{-1}}$ …………………………………… (2)
Now let find the required dimension through dimensional analysis.
$\dfrac{\omega }{k}\propto {{M}^{\alpha }}{{L}^{\beta }}{{T}^{\gamma }}$
$\Rightarrow \dfrac{\omega }{k}=n{{M}^{\alpha }}{{L}^{\beta }}{{T}^{\gamma }}$
Taking dimensions on both sides,
$\Rightarrow \dfrac{\left[ \omega \right]}{\left[ k \right]}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{\beta }}{{\left[ T \right]}^{\gamma }}$
Substituting (1) and (2),
$\Rightarrow \dfrac{{{\left[ T \right]}^{-1}}}{{{\left[ L \right]}^{-1}}}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{\beta }}{{\left[ T \right]}^{\gamma }}$
$\Rightarrow {{\left[ T \right]}^{-1}}{{\left[ L \right]}^{1}}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{\beta }}{{\left[ T \right]}^{\gamma }}$
So by principle of homogeneity of dimensions, we have,
$\alpha =0$
$\beta =1$
$\gamma =-1$
Therefore,
$\Rightarrow \dfrac{\left[ \omega \right]}{\left[ k \right]}={{\left[ M \right]}^{0}}{{\left[ L \right]}^{1}}{{\left[ T \right]}^{-1}}$
So, the dimensions of $\dfrac{\omega }{k}$ are $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$

So, the correct answer is “Option B”.

Note: You could save a lot of time by avoiding the steps involving dimensional analysis. We have solved this question in detail for better understanding. You could actually get the answer direction from the expression for angular frequency and angular wave number. While doing the dimensional analysis we have introduced a dimensionless constant n just for the purpose of converting the proportionality to equation.