A 100 g body hung on a spring, which elongates the spring by 4 cm. When a certain object is hung on the spring and set vibrating, its period is 0.568 s. What is the mass of the object pulling the spring? (Ans: 0.20 kg)


Data Given:

Mass of the body = m₁ = 100 g = 0.1 kg

Elongation = x = 4 cm = 0.04 m

Time period  = T = 0.568 s

Value of acceleration due to gravity = g = 9.8 m s⁻²



To Find:

Mass of the object hanged = m₂ = ?


Solution:


Here we will find the spring constant by using the Hook's law from given data for body of mass m₁ attached to spring

F = Kx

or

mg = Kx

or

K = mgx

by putting values

K = 0.1kgｘ9.8ms⁻²0.04m

K = 24.5 N m⁻¹


Now using the formula for Time Period of mass attached to spring for object of mass m₂ 

T = 2π √m2K

To separate the m₂ from this formula we will take square both side and by simplifying we get

m₂ = T2K(2π)2

by putting values

m₂ = (0.568s)2ｘ24.5Nm⁻¹(2ｘ3.1416)2
 
m₂ = 0.322624s²ｘ24.5Nm⁻¹(6.283185)2

m₂ = 7.904288kg39.478

m₂ = 0.2002 kg

or

m₂ = 200 g ------------Ans.



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