A200 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

236 oscillations

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

statement #3 on andre"s list is false.