Question

In the relation y=r sin(wt-kx) the dimensions of w/k are

Answer 1

The dimensions of $\bold{\frac{\omega}{k} \text { is }\left[L T^{-1}\right]}$

Solution:

The given equation $y=r \sin (\omega t-k x)$ is the equation denoting a wave form.

And thus for a wave form the dimension of $\omega t \text { and } k x$ must be equal to radians.

So according to dimension analysis the quantity should be equated to get the dimension.  

So equating these two quantities.

$\omega t=k x=r a d i a n s$

Radians = m/s.

So from the above equation

$\frac{\omega}{k}=\frac{x}{t}=\frac{m}{s}$

So, the dimension of $\frac{\omega}{k}$ will be equal to the dimension of $\frac{x}{t}$.

So dimension$\frac{\omega}{k}=\left[L T^{-1}\right]$

Where, L refers the length in metre.

T refers the time in seconds.