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 = Cm^(x)k^(y)`, where C is a dimensionless constant. The values of x and y are

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

by putting the dimensions of each quantity both the sides we get `[T^-1]=[M]^x[MT^-2]^y`
now comparing the dimensions of quantities in both sides we get `x+y=0` and `2y=1` `because` `x=-(1)/(2)`,`y=(1)/(2)`