Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Consider a progressive wave whose displacement at distance $x$ and time $t$ is $y=Asin\left(\right. \omega t - k x \left.\right)$ . If $p=\frac{\omega }{k}$ , then the dimensions $p$ are equivalent to:

NTA AbhyasNTA Abhyas 2020

Solution:

$y=Asin \left(\right. \omega t - k x \left.\right)$
As $\left(\omega t - k x\right)$ represents an angle which is dimensionless, therefore
$\left[\omega \right]=\frac{1}{\left(t\right)}=\left[T^{- 1}\right]and \, \left[k\right]=\frac{1}{\left[x\right]}=\left[L^{- 1}\right]$
$\left[\frac{\omega }{k}\right]=\frac{\left[\right] T^{- 1}}{\left[L^{- 1}\right]}=\left[L T^{- 1}\right]=velocity$