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?

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