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The frequency of vibration r of a mass m suspended from a spring of spring constant k is given by a relation of this type $\mathrm{f}=\mathrm{Cm}^{\mathrm{x}} \mathrm{K}^{\mathrm{y}}$; where c is a dimensionless quantity. The value of x and y are
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The correct answer is:
$x=-\frac{1}{2}, y=\frac{1}{2}$
The correct option is $\mathbf{D}$
$x=-\frac{1}{2}, y=\frac{1}{2}$
The frequency of oscillation spring mass system $=\frac{1}{2 \pi}$ $\sqrt{\frac{k}{m}}$
Where, $\mathrm{k}=$ spring constant
and $\mathrm{m}=$ mass with spring
And given frequency, $\mathrm{f}=\mathrm{cm}^x \mathrm{k}^y$
Comparing above two frequency we get
$x=-\frac{1}{2}, y=\frac{1}{2}$
$x=-\frac{1}{2}, y=\frac{1}{2}$
The frequency of oscillation spring mass system $=\frac{1}{2 \pi}$ $\sqrt{\frac{k}{m}}$
Where, $\mathrm{k}=$ spring constant
and $\mathrm{m}=$ mass with spring
And given frequency, $\mathrm{f}=\mathrm{cm}^x \mathrm{k}^y$
Comparing above two frequency we get
$x=-\frac{1}{2}, y=\frac{1}{2}$
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