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31 December, 18:28

A 200 g oscillator in a vacuum chamber has a frequency of 2.0 hz. when air is admitted, the oscillation decreases to 60% of its initial amplitude in 50 s. how many oscillations will have been completed when the amplitude is 30% of its initial value?

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  1. 31 December, 18:30
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    To solve this given problem, we make use of the formula:

    A = Ao e^ ( - b t / 2 m)

    Substituting all the given values into the equation:

    A / Ao = e^ ( - b t / 2 m)

    When A / Ao = 0.60 and t = 50 s, we find for b:

    0.60 = e^ ( - b t / 2 m)

    ln (0.60) = - b t / 2 m

    b = - (2 m) ln (0.60) / t

    b = ( - 2) (.200) ln (0.60) / 50

    b =.00409

    When A / Ao = 0.30, we find for t:

    0.30 = e^ ( - b t / 2 m)

    ln (0.30) = - (0.00409) t / 2 (0.200)

    t = - (0.200) ln (0.30) / 0.00409

    t = 118 s

    Therefore the number of oscillations is:

    oscillations = f * t = 2 s^-1 (118 s) = 236

    Answer: 236 oscillations
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