Question

The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant $k$ is given by a relation $f = am^x\, k^y$, where $a$ is a dimensionless constant. The values 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

Answer: (D)

$f=am^x k^y \, \, \, ....(i)$
Dimensions of frequency $f=[M^0 L^0 T^{-1}]$
Dimensions of constant $a=[M^0 L^0 T^0]$
Dimensions of mass $m = [M]$
Dimensions of spring constant $k = [MT^{-2}]$
Putting these value in equation $(i)$, we get
$[M^0 L^0 T^{-1}]=[M]^x[MT^{-2}]^y$
Applying principle of homogeneity of dimensions, we get
$x+y=0 .....(ii)$
$-2y=-1 ......(iii)$
or $y=\frac{1}{2}, x=-\frac{1}{2}$