Question

If $x=ksin\left(klt\right)$, where $x$ is displacement and $t$ is time. Then dimensional formula for $l$ will be :

$(k,l=constant)$

Answer 1

If $x=k\sin \left(klt\right)...\left(1\right)$

Here, since on LHS we have displacement and on RHS, we have $\sin \theta$ which is dimensionless.

$\therefore$ $x$ and $k$ must be dimensionally equal.

since, $\left[x\right]=\left[L\right]$

and so does $\left[k\right]=\left[x\right]=\left[L\right]......\left(1\right)$

$\therefore$ For $\sin \theta$ to be dimensionless

$\left[klt\right]=\left[M^0{L}^0{T}^0\right]$

$\left[l\right]=\frac{\left[M^0{L}^0{T}^0\right]}{\left[k\right]\left[t\right]}$

from (1),

$\left[l\right]=\frac{\left[M^0{L}^0{T}^0\right]}{\left[L\right]\left[T\right]}$

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