Question

The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation f = am^xk^y, where a is a dimensionless constant. The values of x and y are

(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 1

Correct Option (d) x = -1/2, y = 1/2

Explanation: 

f = am^xk^y,

Dimensions of frequency f = [M⁰L⁰T^-1]

Dimensions of constant a = [M⁰L⁰T⁰]

Dimensions of mass m = [M]

Dimensions of spring constant k = [MT^-2]

Putting these value in equation (i), we get

 [M⁰L⁰T^-1] = [M]^x [MT^-2]^y

Applying principle of homogeneity of dimensions, we get

x + y = 0     ...(i)

-2y = -1       ....(iii)

or y = 1/2, x = -1/2