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Two underdamped oscillators share the same natural frequency omega0. The first has mass m1 and damping coefficient b1; the second has mass m2 and damping coefficient b2. Each is driven by F_ext = F0*cos(omega*t) and the drive frequency is tuned through resonance. If m1 = 4*m2 and b1 = 2*b2, which statement about the driven resonance peaks is correct?

1. The resonant peak of the first oscillator is higher and narrower than that of the second.
2. The resonant peak of the first oscillator is higher and wider than that of the second.
3. The resonant peak of the first oscillator is lower and wider than that of the second.
4. The resonant peak of the first oscillator is lower and narrower than that of the second.

Correct answer: The resonant peak of the first oscillator is higher and narrower than that of the second.

Solution

Driven-oscillator amplitude at resonance is A_max approx F0/(b*omega0), so a smaller b gives a higher peak - but here b1 = 2*b2 (larger), so by amplitude alone the first would be lower. The sharpness is governed by the damping rate gamma = b/m: gamma1 = b1/m1 = 2*b2/(4*m2) = b2/(2*m2) = gamma2/2. A smaller gamma means a higher quality factor Q = omega0/gamma, i.e. a narrower and taller resonance. Since Q1 = 2*Q2, the first oscillator's resonance is both higher and narrower. The width/Q effect (Q1 = 2Q2) dominates, giving a higher and narrower peak for oscillator 1.

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