Question

The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant 𝐾 is given by a relation of this type $f=C{m}^x{K}^y$, where $C$ is a dimensionless quantity. The value of $x$ and $y$ are $P$ and $Q$, respectively.
Find $(P+Q)$

Answer 1

Given:
$f=C{m}^x{K}^y...\left(i\right)$
As we know
Dimension of $\left[f\right]=\left[T^{-1}\right]$
Dimension of $\left[m\right]=\left[M\right]$
Dimension of $\left[K\right]=\left[\frac{ML{T}^{-2}}{L}\right]=\left[M{T}^{-2}\right]$
By putting all dimensions in eqaution (i) we get,
$\left[f\right]={\left[M\right]}^x{\left[M{T}^{-2}\right]}^y$
$\left[M^0{L}^0{T}^{-1}\right]={\left[M\right]}^x{\left[M{T}^{-2}\right]}^y$
By comparing L.H.S= R.H.S
$x+y=0$
$-2y=-1$
$y=\frac{1}{2}$
$x=-\frac{1}{2}$
Hence, $x=-\frac{1}{2},y=\frac{1}{2}$
$\left[f\right]=\left[Cm\frac{-1}{2}K\frac{1}{2}\right]$
As given, the value of $x$ and $y$ are $P$ and $Q$.
Therefore, $P=-\frac{1}{2},Q=\frac{1}{2}$
$(P+Q)=-\frac{1}{2}+\left(\frac{1}{2}\right)=0$