Question

A 199 g oscillator in a vacuum chamber has a frequency of 1.58 Hz . When air is admitted, the oscillation decreases to 44.2 % of its initial amplitude in 46.5 s . What is the dampening constant of the oscillation? The answer needs to be in the unit - seconds.

Answer 1

The dampening constant of the oscillation is 0.00698 kg/s.

Step-by-step explanation:

Given that,

Frequency = 1.58 Hz

Time = 46.5 s

We need to calculate the dampening constant of the oscillation

Using equation of decay of amplitude

`A=A_(0)e^{(-bt)/(2m)}`....(I)

`(A)/(A_(0))=(44.2)/(100)`

Put the value into the equation (I)

`(44.2)/(100)=e^{(-bt)/(2m)}`

`ln((44.2)/(100))=(-bt)/(2m)`

`ln((44.2)/(100))=(-b*46.5)/(2*199*10^(-3))`

`-b=(2*199*10^(-3)ln(0.442))/(46.5)`

`b=0.00698\ kg/s`

Hence, The dampening constant of the oscillation is 0.00698 kg/s.