A body of mass 'm', suspended from a spring with force constant k, vibrates with frequency 'f₁ ' When its length is cut into half and the same body is suspended from one of the halves, the frequency is 'f₂'. Find out f₁ f₂⁻¹. (f₁ f₂⁻¹  = 0.707)




Given:

Mass of the body = m
Spring constant = k
Frequency of the vibration = f₁

Let the initial displacement of the suspended spring = x₁
After reducing to half the displacement = x₂ = x₁/2
Frequency after reducing length = f₂


To Find:

f₁ f₂⁻¹ = f₁f₂ = ?



Solution:


The general formula for the Time Period of a mass suspended from a spring is 

T = 2π √mk

but f = 1T and K = Fx =  mgx

by putting these value 

f = 12π√mmgx


f = 12π √xg



Now for f₁

f₁ = 12π √x₁g ----------(1)

and for f₂

f₂ = 12π √x₂g 

putting x₂ = x₁/2

f₂ = 12π √x₁2g ------------(2)


Dividing equation (1) by eqn (2) the quantities 12π, x₁ and g will cancel and we get


f₁f₂ = √12


f₁f₂ = √0.5


f₁f₂ = 0.707

or

f₁ f₂⁻¹ = 0.707 --------------Ans.



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