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

• A. $x=\frac{1}{2},y=\frac{1}{2}$
• B. $x=-\frac{1}{2},y=-\frac{1}{2}$
• C. $x=\frac{1}{2},y=-\frac{1}{2}$
• D. $x=-\frac{1}{2},y=\frac{1}{2}$

Answer 1

The correct option is D $x=-\frac{1}{2},y=\frac{1}{2}$

By putting the dimensions of each quantity in both the sides we get $\left[T^{-1}\right]=\left[M{]}^x\right[M{T}^{-2}{]}^y$
$\left[M^0{L}^0{T}^{-1}\right]=\left[M^{x+y}{L}^0{T}^{-2y}\right]$
Now, comparing the dimensions of quantities in both sides we get $x+y=0$ and $2y=1$
$\therefore x=-\frac{1}{2},y=\frac{1}{2}$