Question

The equation of wave is given by $Y=A\sin \omega \left( \dfrac{x}{v}-k \right)$ where $\omega $ is the angular velocity and v is the linear velocity. The dimension of k is:
$\left( a \right)\left[ LT \right]$
$\left( b \right)\left[ T \right]$
$\left( c \right)\left[ {{T}^{-1}} \right]$
$\left( d \right)\left[ {{T}^{-2}} \right]$

Answer 1

Hint: As the question has asked about the dimension of k which is in the wave equation so, we will keep the two terms present inside the brackets of the equation in order to find the required dimension. This is because these two will carry the same dimension as being inside the same sine angle after that substitute values of dimensions as required.
Formula used:
$\text{Dimension}\left( \dfrac{x}{v} \right)=\text{Dimension}\left( k \right)$ where $\omega $ is the angular velocity and v is the linear velocity and k is a variable.

Complete answer:
Angular velocity: We use angular velocity to find the velocity of any object in circular motion. This type of velocity depends upon direction as well as magnitude. The magnitude of such velocity is the same as the speed of the object while its direction is always taken to be normal to its own plane in a circular motion.
We can always compare the angular velocity to linear velocity. This totally depends upon the equation defining the relation between these two terms. The equation is $v=r\omega $ .

Dimensions: One can find the very basic dimensions in physics. These are time, length and mass. These are denoted by t, l and m respectively.

Now, we will consider the equation of wave $Y=A\sin \omega \left( \dfrac{x}{v}-k \right)$. Since, we can see that the terms inside the bracket come under algebra so, these will process the same dimension formula.

As x will work as dimension L only and the dimensions for velocity v is $\left[ L{{T}^{-1}} \right]$ so, we will get that,

$\begin{align}
  & \text{Dimension}\left( \dfrac{x}{v} \right)=\text{Dimension}\left( k \right) \\
 & \Rightarrow \text{Dimension}\left( k \right)=\dfrac{\left[ L \right]}{\left[ L{{T}^{-1}} \right]} \\
 & \Rightarrow \text{Dimension}\left( k \right)=\dfrac{1}{\left[ {{T}^{-1}} \right]} \\
 & \Rightarrow \text{Dimension}\left( k \right)=\left[ {{T}^{-1}} \right] \\
\end{align}$

So, the correct answer is “Option B”.

Note:
This question can only be done by using the trick of placing two terms inside the brackets of sine angle. Initially, it seems to be a little bit confusing but once the trick is understood then it becomes easy to solve. If we need to solve the question correctly then we must be aware of the values of dimensions which are used in finding that of k. The angular velocity is taken normal to the circular motion on the other hand the linear one is not taken like this. We should always remember to put brackets while writing dimensions otherwise, the answer will not be accepted.