Q. The frequency $f$ of vibration of mass $m$ suspended from a spring of spring constant $k$ is given by $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f = cm^{x}k^{y}$ Where $c$ is dimensionless constant. The values of $x\ and\ y$ are respectively
A) 1/2, 1/2
B) -1/2, 1/2
C) 1/2, -1/2
D) -1/2, -1/2
Explanation
Frequency $f = cm^{x}k^{y},\ k = Force/Length$ $\left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack = \lbrack M\rbrack^{x}\left\lbrack ML^{0}T^{- 2} \right\rbrack^{y}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \lbrack M\rbrack^{x + y}\ \lbrack L\rbrack^{0}\lbrack T\rbrack^{- 2y}$ Comparing the powers on M, L and T $\ \ \ \ \ \ \ \ \ \ \ \ \ - 2y = - 1$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ y = \frac{1}{2}$ And $x + y = 0$ $\therefore\ \ x = \ - y = - \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.