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Question

For the damped oscillator system shown with m = 250g, k = 85 N/m, and b = 70 g/s, what is the ratio of the oscillation amplitude at the end of 20 cycles to the initial oscillation amplitude?


A

1:2

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B

20:21

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C

19:50

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D

5:6

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Solution

The correct option is C

19:50


The angular frequency of oscillation in this case will be equal to

kmb24m2

ω=850.25(0.07)24×(0.25)2340=18.4 rad s1

T=2πω=0.34 s

In 20 cycles 20 time periods must have passed, so time after 20 cycles, t=20×0.34

=6.8 s

We know A=A0ebt2m

A=A0e(0.07×6.82×0.25)=0.38 A0

So the ratio A:A0 = 0.38:1 or 19:50


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