Question

$\omega t=\theta$

$Kx=\theta$

What will be dimensions unit of $K$ in the above formula. Given $x$ is displacement and $\theta$ is angular displacement and $\omega$ is angular frequency.

Answer 1

Given that,

$\omega t=\theta$

$Kx=\theta$

distance $=x$

Now,

The dimension of $\omega$

$\omega t=\theta$

$\omega =\frac{\theta }{t}$

$\omega =\frac{1}{T}$

$\omega =T^{-1}$

$\omega =\left[M^0{L}^0{T}^{-1}\right]$

Now, for $Kx=\theta$

The dimension of $K$

$Kx=\theta$

$K=\frac{\theta }{x}$

$\omega =\frac{1}{l}$

$K=L^{-1}$

$K=\left[M^0{L}^{-1}{T}^0\right]$

Hence, this is the required solution