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 / 2.5 m)
ln ( 0.60 ) = − b t / 2.5 m
b = − ( 2.5 m ) ln ( 0.60 ) / t
b = ( − 2 ) ( 0.200 ) ln ( 0.60 ) / 50
b = .004086
When A / Ao = 0.30, we find for t:
0.30 = e^( − b t / 2.5 m)
ln ( 0.30 ) = − (0.004086) t / 2 (0.210)
t = − (0.210) ln ( 0.30 ) / 0.004086
t = 61.878 s
Therefore the number of oscillations is:
oscillations = f * t = 2 s^-1 (61.878 s) = 123.756
Review l Constants l Periodic Table Part A A 210 g oscillator in a vacuum chamber...
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