In the relation \(y=r \sin (\omega t-k x)\), the dimensions of \(\omega / k\) are
In the relation \(y=r \sin (\omega t-k x)\), the dimensions of \(\omega / k\) are
- \(\left[M^{0} L^{0} T^{0}\right]\)
- \(\left[M^{0} L^{1} T^{-1}\right]\)
- \(\left[M^{0} L^{0} T^{1}\right]\)
- \(\left[M^{0} L^{\prime} T^{0}\right]\)
Solution
$\begin{aligned}
&\text{b. } y=r \sin (\omega t-k x) \\
&\text{Here, } \omega t=\text{angle } \Rightarrow \quad \omega=\frac{1}{T}=T^{-1} \\
&\text{Similarly, } k x=\text{angle } \Rightarrow \quad k=\frac{1}{x}=L^{-1} \\
&\therefore \quad \frac{\omega}{k}=\frac{T^{-1}}{L^{-1}}=L T^{-1}
\end{aligned}$
Or simply \(\frac{\omega}{k}\) represents wave velocity \(\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda=v\), where \(f\) is frequency
Asked in: JEE Mains - Units and Dimensions - Chapter Test
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