Question
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 constant. The values 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.
$$x = - \frac{1}{2},y = \frac{1}{2}$$
Answer :
$$x = - \frac{1}{2},y = \frac{1}{2}$$
Solution :
$$\eqalign{
& {\text{As}}\,f = C{m^x}{k^y} \cr
& \therefore \left( {{\text{Dimension of}}\,f} \right) = C{\left( {{\text{dimension of }}m} \right)^x} \times {\left( {{\text{dimensions of }}k} \right)^y} \cr
& \left[ {{T^{ - 1}}} \right] = C{\left[ M \right]^x}{\left[ {M{T^{ - 2}}} \right]^y}\,......\left( {\text{i}} \right)\,\left( {{\text{where,}}\,k = \frac{{{\text{force}}}}{{{\text{length}}}}} \right) \cr} $$
Applying the principle of homogeneity of dimensions, we get
$$\eqalign{
& x + y = 0, - 2y = - 1\,{\text{or}}\,y = \frac{1}{2} \cr
& \therefore x = - \frac{1}{2} \cr} $$