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The angular frequency of a damped oscillator is omega = sqrt(k/m - r²/(4*m²)), where k is the spring constant, m the mass, and r the damping constant. If r²/(m*k) = 0.08 (8%), find the approximate change in time period compared to the undamped oscillator.

1. decreases by 8%
2. decreases by 1%
3. increases by 1%
4. increases by 8%

Correct answer: increases by 1%

Solution

omega = sqrt(k/m)*sqrt(1 - r²/(4mk)) = omega0*(1 - r²/(8mk)) for small damping. With r²/(mk) = 0.08, the fractional decrease in omega is r²/(8mk) = 0.08/8 = 0.01 = 1%. Since T = 2*pi/omega, T increases by about 1%.

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