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

The Correct Option is 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 $y=\frac{1}{2}, x=-\frac{1}{2}$