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?

Respuesta :

To solve this given problem, we make use of the formula:

A = Ao e^( βˆ’ b t / 2 m)

Substituting all the given values into the equation:

A / Ao = e^( βˆ’ b t / 2 m)

Β 

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

0.60 = e^( βˆ’ b t / 2 m)

ln ( 0.60 ) = βˆ’ b t / 2 m

b = βˆ’ ( 2 m ) ln ( 0.60 ) / t

b = ( βˆ’ 2 ) ( .200 ) ln ( 0.60 ) / 50

b = .00409

Β 

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

0.30 = e^( βˆ’ b t / 2 m)

ln ( 0.30 ) = βˆ’ (0.00409) t / 2 (0.200)

t = βˆ’ (0.200) ln ( 0.30 ) / 0.00409

t = 118 s

Β 

Therefore the number of oscillations is:

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

Β 

Answer: 236 oscillations

Β