Frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a rela

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1.Units, Dimensions and Measurement

medium

The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant $K$ is given by a relation of this type $f = C\,{m^x}{K^y}$; where $C$ is a dimensionless quantity. The value 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

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

(AIPMT-1990)

Solution

(d) By putting the dimensions of each quantity both the sides we get $[{T^{ – 1}}] = {[M]^x}{[M{T^{ – 2}}]^y}$

Now comparing the dimensions of quantities in both sides we get $x + y = 0\;{\rm{and }}\,2y = 1$

$\therefore $ $x = – \frac{1}{2},\,\,y = \frac{1}{2}$

Standard 11

Physics

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