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| | | ### Statement |
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| | − | $3.5.38.$ The figure shows the dependence of the square of the amplitude of the velocity of forced oscillations on the frequency of the forcing force, the amplitude of which is constant. Determine the natural frequency of the oscillator, its damping coefficient and quality factor. |
| | + | 3.5.38. The figure shows the dependence of the square of the amplitude of the velocity of forced oscillations on the frequency of the forcing force, the amplitude of which is constant. Determine the natural frequency of the oscillator, its damping coefficient and quality factor. |
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| | | ### Solution |
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| | − |  |
| | + | The energy (and hence the square of the amplitude of the velocity) of a weakly-damped forced oscillator, where $\gamma \ll \omega_0$, near resonance is proportional to: |
| | + | $$\frac{1}{(\omega - \omega_0)^2 + \gamma^2}$$ |
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| | − | The energy (and hence the square of the amplitude of the velocity) of a weakly-damped forced oscillator, where $\gamma\ll\omega_0$, near resonance is proportional to |
| | + | This means that the resonance curve peaks at $\omega = \omega_0$ and the width of the curve at half of the maximum height is $2\gamma$. |
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| | − | \[\frac{1}{(\omega-\omega_0)^2+\gamma^2}.\] |
| | + | Reading from the provided graph, the peak is symmetric and located exactly halfway between the 500 s⁻¹ and 600 s⁻¹ marks, which gives the natural frequency: |
| | + | $$\omega_0 = 550 \text{ s}^{-1}$$ |
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| | − | This means that the resonance curve peaks at $\omega=\omega_0$ and the width of the curve at half of the maximum height is $2\gamma$. Reading from the provided graph, we have $\omega_0\approx540$ s$^{-1}$ and $\gamma\approx60$ s$^{-1}$, and the quality factor is $Q=\omega_0/(2\gamma)\approx4.5$. |
| | + | The red horizontal line indicates half of the maximum height. The corresponding vertical red lines intersect the frequency axis exactly at 500 s⁻¹ and 600 s⁻¹. Therefore, the width of the curve at half-maximum is: |
| | + | $$2\gamma = 600 - 500 = 100 \text{ s}^{-1} \implies \gamma = 50 \text{ s}^{-1}$$ |
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| | − | #### Answer |
| | + | The quality factor is calculated as: |
| | + | $$Q = \frac{\omega_0}{2\gamma} = \frac{550}{100} = 5.5$$ |
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| | − | $\omega_0\approx540$ s$^{-1}$, $\gamma\approx60$ s$^{-1}$, $Q\approx4.5$ |
| | + | #### Answer |
| | + | $$\omega_0 = 550 \text{ s}^{-1}, \gamma = 50 \text{ s}^{-1}, Q = 5.5$$ |
