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16 December, 10:33

What is the frequency of a simple pendulum 2.7 m long in the following situations?

(a) in a room that is stationary Hz

(b) in an elevator accelerating upward at a rate of 1.1 m/s2 Hz

(c) in free fall Hz

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  1. 16 December, 12:14
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    Answer: (a) 0.35Hz (b) 0.37Hz (c) 0

    Explanation: frequency of a simple pendulum is expressed as;

    f = 1/2π√ (a/L)

    (a) in a room that is stationary, a, is the acceleration due to gravity = 9.8m/s²

    f = 1/2π√ (g/L)

    f = 1/2π (9.8/2)

    f = 0.35Hz

    (b) If the pendulum is in an elevator accelerating at a rate of 1.1m/s²

    The total force will surely to both a and g

    Therefore apparent acceleration = a+g = 1.1 + 9.8=10.9 m/s²

    f = 1/2π √ (10.9/2)

    f = 0.37Hz

    (c) For a free fall, apparent acceleration is 0

    f = 1/2π √ (0/L)

    f = 0
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