Q. In the relation $y = r$ sin $(\omega t - kx),\$the dimensions of $\omega/k$ are
A) $\lbrack M^{0}L^{0}T^{0}\rbrack$
B) $\lbrack M^{0}L^{1}T^{- 1}\rbrack$
C) $\lbrack M^{0}L^{0}T^{1}\rbrack$
D) $\lbrack M^{0}L^{1}T^{0}\rbrack$
Explanation
$y = r$ sin $(\omega t - kx)$ There $\omega t =$angle $\therefore\ \ \ \ \omega = \frac{1}{T} = T^{- 1}$ Similarly $kx =$angle $\therefore\ \ \ k = \frac{1}{x} = L^{- 1}$ $\therefore\ \ \ \frac{\omega}{k} = \frac{\lbrack T^{- 1}\rbrack}{\left\lbrack L^{- 1} \right\rbrack} = \lbrack LT^{- 1}\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.