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24 November, 19:29

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

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  1. 24 November, 21:20
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    The add mass = 5.465 kg

    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.

    Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

    Substituting into equation 4

    m₂ = (2.07) ² (0.5) / (1.18) ²

    m₂ = 4.285 (1.392)

    m₂ = 5.965 kg.

    Added mass = m₂ - m₁

    Added mass = 5.965 - 0.5

    Added mass = 5.465 kg.

    Thus the add mass = 5.465 kg
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