00:01
For part a, we know that the maximum displacement for the damped oscillator, this would be equalling to the amplitude multiplied by e raised to the negative t over two times tau, the time constant.
00:16
We can then say negative time over two times tau, the time constant equals the natural log of the maximum position as a function of time.
00:27
This would be divided by the amplitude a.
00:31
And so we can solve the time constant is then equalling to negative 0 .50 seconds divided by 2 times the natural log of 0 .98.
00:49
This is equaling then to 12 .375 seconds.
00:56
And so we can say that then 25 oscillations will be completed in t equaling 25 multiplied by t, and this is equaling to 12 .5 seconds.
01:13
And so at that time, we can say that the amplitude or the maximum exposition at t equaling 12 .5 seconds, this would be equaling to 10 centimeters, multiplied by e, raised to the negative 12 .5 seconds divided by 2 multiplied by 12 .375 seconds...