The displacement of a progressive wave is represented by $y = A\,sin \,(\omega t - kx)$ where $x$ is distance and t is time. Write the dimensional formula of $(i)$ $\omega $ and $(ii)$ $k$.
Now, $[\mathrm{LHS}]=[\mathrm{RHS}]$
$[y]=[\mathrm{A}]=\mathrm{L}$
because $\omega t-k x$ is dimensionless,
$[k x]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\therefore[\omega t]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\therefore [k] \mathrm{L}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\therefore[\omega] \mathrm{T}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$
$\therefore [k]=\mathrm{L}^{-1}\therefore[\omega]=\mathrm{T}^{-1}$
If speed $V,$ area $A$ and force $F$ are chosen as fundamental units, then the dimension of Young's modulus will be :
$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length $(l)$, mass of the bob $(m)$ and acceleration due to gravity $(g)$. Derive the expression for its time period using method of dimensions.
A force defined by $F=\alpha t^2+\beta t$ acts on a particle at a given time $t$. The factor which is dimensionless, if $\alpha$ and $\beta$ are constants, is:
If force $[F],$ acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of energy.