The frequency of vibration f of a mass m susp | vedclass

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(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.The dimensional formula fo

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$x = -\frac{1}{2}, y = \frac{1}{2}$

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(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.The dimensional formula for spring constant $K$ is $[M^1 L^0 T^{-2}]$.Given the relation $f = C m^x K^y$,substituting the dimensions on both sides:$[M^0 L^0 T^{-1}] = [M^1]^x [M^1 T^{-2}]^y$$[M^0 L^0 T^{-1}] = [M^{x+y} T^{-2y}]$Comparing the powers of $M$ and $T$ on both sides:For $M$: $x + y = 0$For $T$: $-2y = -1 \implies y = \frac{1}{2}$Substituting $y = \frac{1}{2}$ into $x + y = 0$,we get $x + \frac{1}{2} = 0 \implies x = -\frac{1}{2}$.Thus,$x = -\frac{1}{2}$ and $y = \frac{1}{2}$.

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 quantity. The values of $x$ and $y$ are:

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