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

(4) `f = C m^(x) k^(y)`
`T^(-1) = [M]^(x) [MT^(-2)]^(y)`
`M^(0) L^(0) T^(-1) = M^(x + y) T^(-2y)`
Comparing powers of `M` and `T`
`M : x + y = 0`
`T : - 2y = - 1 implies (1)/(2), x = - (1)/(2)`
Or
Time-period simple pendulam
`T = 2 pi sqrt(m//k)`
`f = (1)/(T) = (1)/(2pi) ((k)/(m))^(1//2)`
`f = C m^(x) k^(y)`
Comparing powers of `m` and `k`
`x = (1)/(2), y = (1)/(2)`