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11 September, 17:09

Two simple pendulum of slightly different length, are set off oscillating in step is a time of 20s has elasped, during which time the longer pendulum has completed exactly 10 oscillations. Find the length of each pendulum.

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  1. 11 September, 17:13
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    Length of longer pendulum = 99.3 cm

    Length of shorter pendulum = 82.2 cm

    Explanation:

    Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

    From T = 2π√ (l/g), the length of the pendulum. l = T²g/4π²

    substituting T = 2s and g = 9.8 m/s² we have

    l = T²g/4π²

    = (2 s) ² * 9.8 m/s² : 4π²

    = 39.2 m : 4π²

    = 0.993 m

    = 99.3 cm

    Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

    So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

    t/T₂ = t/T₁ + x

    1/T₂ = 1/T₁ + x (1)

    Also, the T₂ = t/n₂ = t / (n₁ + x) (2)

    From (1) T₂ = T₁ / (T₁ + x) (3)

    equating (2) and (3) we have

    t / (n₁ + x) = T₁ / (T₁ + x)

    substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

    20 s / (10 + x) = 2 / (2 + x)

    10 / (10 + x) = 1 / (2 + x)

    (10 + x) / 10 = (2 + x)

    (10 + x) = 10 (2 + x)

    10 + x = 20 + 10x

    collecting like terms

    10x - x = 20 - 10

    9x = 10

    x = 10/9

    x = 1.11

    x ≅ 1 oscillation

    substituting x into (2)

    T₂ = t/n₂ = t / (n₁ + x)

    = 20 / (10 + 1)

    = 20/11

    = 1.82 s

    Since length l = T²g/4π²

    substituting T = 1.82 s and g = 9.8 m/s² we have

    l = T²g/4π²

    = (1.82 s) ² * 9.8 m/s² : 4π²

    = 32.46 m : 4π²

    = 0.822 m

    = 82.2 cm
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