The frequency `f` of vibrations of a mass `m` suspended from a spring of spring constant `k` is given by `f = Cm^(x) k^(y)` , where `C` is a dimensionnless constant. The values of `x and y` are, respectively,

Options

A. `x = (1)/(2),y = (1)/(2)`
B. `x =- (1)/(2), y =- (1)/(2)`
C. `x = (1)/(2), y =- (1)/(2)`
D. `x =- (1)/(2), y =(1)/(2)`

Answer

D

Explanation

As `f = Cm^(x) k^(y)`
`:. `(Dimension of f) `= C ("dimension of" m)^(x) xx ("dimension of " k)^(y)`
`[T^(-)] = C[M]^(x) [MT^(-2)]^(y)` ...(i)
(where, `k = ("force")/("length")`)
Applying the principle of homogeneity of dimensions, we get
`x + y = 0, -2y =- 1` or `y = (1)/(2)`
`:. x =- (1)/(2)`