The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant $k$ is given by relation of the type $f = cm^{x}k^{y}$, where $c$ is a dimensionless constant. The values of $x$ and $y$ are

Question

Dr. Abubacker · Accepted Answer

$\left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack = \left\lbrack M^{x} \right\rbrack\left\lbrack M^{y}T^{2y} \right\rbrack = \lbrack M^{x + y}T^{- 2y}\rbrack$ Equating powers of M and T. $x + y = 0, - 2y = - 1$ Or $y = \frac{1}{2},x + \frac{1}{2} = 0$ or $x = - \frac{1}{2}$

Q. The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant $k$ is given by relation of the type $f = cm^{x}k^{y}$, where $c$ is a dimensionless constant. The values of $x$ and $y$ are

Explanation

Reviewed by Dr. Abubacker

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