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 = `\frac {f₁}{f₂}` = ?



Solution:


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

T = 2π `\sqrt frac{m}{k}`

but f = `\frac {1}{T}` and K = `\frac {F}{x}` =  `\frac {mg}{x}`

by putting these value 

f = `\frac {1}{2π sqrt frac{m}{frac {mg}{x}}}`


f = `\frac {1}{2π}` `\sqrt frac{x}{g}`



Now for f₁

f₁ = `\frac {1}{2π}` `\sqrt frac{x₁}{g}` ----------(1)

and for f₂

f₂ = `\frac {1}{2π}` `\sqrt frac{x₂}{g}` 

putting x₂ = x₁/2

f₂ = `\frac {1}{2π}` `\sqrt frac{x₁}{2g}` ------------(2)


Dividing equation (1) by eqn (2) the quantities `\frac {1}{2π}`, x₁ and g will cancel and we get


`\frac {f₁ }{f₂}` = `\sqrt frac{1}{2}`


`\frac {f₁ }{f₂}` = `\sqrt {0.5}`


`\frac {f₁ }{f₂}` = 0.707

or

f₁ f₂⁻¹ = 0.707 --------------Ans.



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