MEDIUM · DAMPED AND DRIVEN OSCILLATIONS

Q FACTOR FROM BANDWIDTH

A tuning fork vibrates at the musical note A₄ (440 Hz), modelled as an underdamped oscillator with natural angular frequency ω₀ = 2π × 440 rad/s and damping coefficient γ = 8.796 rad/s. (a) Compute the quality factor Q. (b) Find how many full oscillation cycles pass before the amplitude falls to 1/e of its initial value. (c) State the half-power bandwidth Δω of this resonator.

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Step-by-step solution

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Step 1

Calculate the quality factor Q = ω₀ / γ.

Hint

Q is a dimensionless ratio comparing the oscillation frequency to the damping rate. A higher Q means the oscillator rings for longer.

Step 2

Find the time t_e at which the amplitude first drops to A₀/e.

Step 3

State the half-power bandwidth Δω of this resonance peak.

Solution walkthrough
The quality factor Q = ω₀/γ = 2π × 440 / 8.796 ≈ 314. This matches a good tuning fork — it rings for roughly a few seconds at 440 Hz. The amplitude envelope is A₀ e^(−γt/2). Setting e^(−γt_e/2) = e^(−1) gives t_e = 2/γ ≈ 0.227 s. In that time the fork completes n_e = t_e × f_d ≈ 0.227 × 440 ≈ 99.9 ≈ 100 cycles, in agreement with the rule of thumb Q/π. The half-power bandwidth of the resonance is Δω = γ = 8.796 rad/s; the resonance peak sits at ω₀ and falls to half its peak power at ω₀ ± γ/2. The two most common errors in Q calculations are using the wrong factor of 2 (writing Q = ω₀/(2γ)) and computing bandwidth in Hz without converting from rad/s. Remember: Δω in rad/s equals γ; the corresponding bandwidth in Hz is Δf = γ/(2π).
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