Answer:
<em> The add mass = 5.465 kg</em>
Explanation:
Note: Since the spring is the same, the length and Tension are constant.
f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)
Where f = frequency, m = mass of the spring.
But f = 1/T ..................... Equation 2
Substituting Equation 2 into equation 1.
1/T ∝ √(1/m)
Therefore,
T ∝ √(m)
Therefore,
T₁/√m₁ = k
where k = Constant of proportionality.
T₁/√m₁ = T₂/√m₂ ........................ Equation 3
making m₂ the subject of the equation
m₂ = T₂²(m₁)/T₁²........................... Equation 4
Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.
<em>Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.</em>
<em>Substituting into equation 4</em>
<em>m₂ = (2.07)²(0.5)/(1.18)²</em>
<em>m₂ = 4.285(1.392)</em>
<em>m₂ = 5.965 kg.</em>
Added mass = m₂ - m₁
Added mass = 5.965 - 0.5
Added mass = 5.465 kg.
<em>Thus the add mass = 5.465 kg</em>
