Question

Review l Constants l Periodic Table Part A A 210 g oscillator in a vacuum chamber has a frequency of 2.5 Hz. When air is admitted, the oscillation decreases to 60% of its initial amplitude in 50 s How many oso lations will have been completed when the amplitude is 30% of its initial value? Express your answer in oscillations ΑΣ oscillations Submit Request Anseer

Answer 1

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