Question

The frequency of vibration $f$ of a mass m suspended from a spring of spring constant $k$ is given by a relation of the type $f=c \, m^{x} \, k^{y},$ where $c$ 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)

$f=c m^{x} k^{y}$
Spring constant $k=$ force/length.
${\left[M^{0} L^{0} T^{-1}\right]=\left[M^{x}\left((M T)^{-2}\right)^{y}\right]=\left[M^{x+y} T^{-2 y}\right]} $
$\Rightarrow x+y=0,-2 y=-1 $ or $ y=\frac{1}{2}$
Therefore, $x=-\frac{1}{2}$