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We know that if the driving frequency \(\omega\) is varied, the maximum response \(\left(A^{2}\right)\) of a driven damped oscillator occurs at \(\omega \approx \omega_{\mathrm{o}}\) (if the natural frequency is \(\omega_{\mathrm{o}}\) and the damping constant \(\beta \ll\) \(\omega_{\mathrm{o}}\) ). Show that \(A^{2}\) is equal to half its maximum value when \(\omega \approx \omega_{\mathrm{o}} \pm \beta,\) so that the full width at half maximum is just \(2 \beta\). [Hint: Be careful with your approximations. For instance, it's fine to say \(\left.\omega+\omega_{\mathrm{o}} \approx 2 \omega_{\mathrm{o}}, \text { but you certainly mustn't say } \omega-\omega_{\mathrm{o}} \approx 0 .\right]\)

Short Answer

Expert verified
The full width at half maximum is \(2\beta\).

Step by step solution

01

Understand the Given Relationships

We need to consider the driven damped harmonic oscillator, where the squared amplitude, \( A^2 \), of the oscillator is maximized around the natural frequency \( \omega_o \). The damping constant \( \beta \) is much smaller than \( \omega_o \), \( \beta \ll \omega_o \). We aim to find the condition where \( A^2 \) is half of its maximum value when \( \omega \approx \omega_o \pm \beta \).
02

Recall Formula for Amplitude Squared

The squared amplitude \( A^2 \) of a driven damped oscillator can be expressed as: \[A^2 = \frac{F_0^2}{m^2 \left( (\omega_o^2 - \omega^2)^2 + (2\beta \omega)^2 \right)}\]Here, \( F_0 \) is the driving force amplitude, \( m \) is the mass of the oscillator, \( \omega \) is the driving frequency, \( \omega_o \) is the natural frequency, and \( \beta \) is the damping parameter.
03

Maximum Amplitude Condition

The maximum of \( A^2 \) occurs at \( \omega = \omega_o \) because the term \((\omega_o^2 - \omega^2)^2\) is minimized. Thus, the maximum amplitude is:\[A_{max}^2 = \frac{F_0^2}{m^2 (2\beta \omega_o)^2}\]
04

Half Maximum Amplitude Condition

To find the condition where \( A^2 \) is half of its maximum, set:\[\frac{A^2}{A_{max}^2} = \frac{1}{2}\]
05

Simplify and Solve for Frequency Deviation

Using the formula for \( A^2 \) and the condition for half maximum:\[\frac{F_0^2}{m^2 \left( (\omega_o^2 - \omega^2)^2 + (2\beta \omega)^2 \right)} = \frac{1}{2} \cdot \frac{F_0^2}{m^2 (2\beta \omega_o)^2}\]This implies:\[(\omega_o^2 - \omega^2)^2 + (2\beta \omega)^2 = 2 \cdot (2\beta \omega_o)^2\]Simplifying, consider small \( \omega - \omega_o \) and expand \( \omega^2 \approx \omega_o^2 + 2\omega_o(\omega - \omega_o) \), leading to: \[2\omega_o(\omega - \omega_o) \approx \pm 2\beta \omega_o\]
06

Conclusion for Width at Half Maximum

Thus, \( \omega - \omega_o \approx \pm \beta \). Therefore, the full width at half maximum of \( A^2 \) is \(2\beta\), since these deviations occur symmetrically about \( \omega_o \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude Resonance
Amplitude resonance is a key feature of a driven damped oscillator. It occurs when the system's response (amplitude) is maximized. This is often at a specific driving frequency that matches the natural frequency of the oscillator. When the driving frequency is equal to this so-called resonance frequency, the oscillator reaches its largest amplitude.

In mathematical terms, the squared amplitude \( A^2 \) is maximized when the driving frequency \( \omega \) aligns with the natural frequency \( \omega_o \). During resonance, frictional forces are counteracted perfectly by the driving force, and energy is transferred most efficiently to the system. The system's ability to resonate is what makes tuning forks, musical instruments, and even bridges vibrate at certain frequencies.
Natural Frequency
Natural frequency, symbolized as \( \omega_o \), is the frequency at which a system oscillates when it is not driven by an external source. For mechanical oscillators, this depends on factors like mass and stiffness. In the case of our driven damped oscillator, this is the frequency where, without any external force or damping, the system would oscillate naturally.

A natural frequency is important because it indicates where amplitude resonance will likely occur. If the driving force's frequency matches this natural frequency, the system will resonate strongly, assuming damping is not too strong. As damping increases, however, the system's resonant response is dampened, and the peak of resonance becomes less sharp.
Damping Constant
The damping constant, represented as \( \beta \), describes how quickly oscillations reduce in amplitude over time. In a driven damped oscillator, damping affects how narrow or broad the resonance peak is.

This constant plays a critical role in determining the full width at half maximum (FWHM) of the resonance. The smaller the damping constant, the sharper the resonance peak will be. For a damping constant much smaller than the natural frequency (\( \beta \ll \omega_o \)), the oscillator demonstrates a prominent resonance peak, indicating efficient energy transfer. However, as \( \beta \) increases, this peak flattens, indicating more energy is lost per cycle.
Frequency Deviation
Frequency deviation in a driven damped oscillator refers to the difference between the natural frequency and the driving frequency when the system is at half of its maximum amplitude.

In the exercise, the task was to show that this "half maximum condition" occurs when the driving frequency \( \omega \) is approximately \( \omega_o \pm \beta \). This means that the drive frequency can vary slightly above or below the natural frequency—within the range of the damping constant—before the amplitude significantly reduces.

The frequency deviation contributing to a measurable effect is what signifies the sensitivity of the system to changes in driving frequencies near its natural frequency. This effect is found in various physical systems, making it pivotal to understanding how oscillatory systems behave with different driving forces.

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Most popular questions from this chapter

Write down the potential energy \(U(\phi)\) of a simple pendulum (mass \(m,\) length \(l\) ) in terms of the angle \(\phi\) between the pendulum and the vertical. (Choose the zero of \(U\) at the bottom.) Show that, for small angles, \(U\) has the Hooke's law form \(U(\phi)=\frac{1}{2} k \phi^{2},\) in terms of the coordinate \(\phi .\) What is \(k ?\)

A massless spring is hanging vertically and unloaded, from the ceiling. A mass is attached to the bottom end and released. How close to its final resting position is the mass after 1 second, given that it finally comes to rest 0.5 meters below the point of release and that the motion is critically damped?

The potential energy of a one-dimensional mass \(m\) at a distance \(r\) from the origin is $$U(r)=U_{0}\left(\frac{r}{R}+\lambda^{2} \frac{R}{r}\right)$$ for \(0 < r < \infty,\) with \(U_{\mathrm{o}}, R,\) and \(\lambda\) all positive constants. Find the equilibrium position \(r_{\mathrm{o}} .\) Let \(x\) be the distance from equilibrium and show that, for small \(x\), the PE has the form \(U=\) const \(+\frac{1}{2} k x^{2}\). What is the angular frequency of small oscillations?

Verify that the decay parameter \(\beta-\sqrt{\beta^{2}-\omega_{0}^{2}}\) for an overdamped oscillator \(\left(\beta > \omega_{\mathrm{o}}\right) d e\) creases with increasing \(\beta\). Sketch its behavior for \(\omega_{0} < \beta < \infty\).

(a) If a mass \(m=0.2 \mathrm{kg}\) is tied to one end of a spring whose force constant \(k=80 \mathrm{N} / \mathrm{m}\) and whose other end is held fixed, what are the angular frequency \(\omega\), the frequency \(f\), and the period \(\tau\) of its oscillations? (b) If the initial position and velocity are \(x_{\mathrm{o}}=0\) and \(v_{\mathrm{o}}=40 \mathrm{m} / \mathrm{s},\) what are the constants \(A\) and \(\delta\) in the expression \(x(t)=A \cos (\omega t-\delta) ?\)

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