Write down the solution to the oscillator equation for the case \(\omega_{0}>\gamma\) if the oscillator starts from \(x=0\) with velocity \(v\). Show that, as \(\omega_{0}\) is reduced to the critical value \(\gamma\), the solution tends to the corresponding solution for the critically damped oscillator.

Short Answer

Expert verified
In this problem, we are given an oscillator equation with a damping parameter 𝜸 and the initial condition x=0 with an initial velocity v. The equation for the damped harmonic oscillator is: \( \ddot{x} + 2\gamma\dot{x} + \omega_{0}^2x = 0 \) We find the solution for this equation given that \( \omega_0 > \gamma \) and show that upon reducing \(\omega_{0}\) to the critical value of \(\gamma\), the solution tends to the corresponding solution for the critically damped oscillator. The general solution for an underdamped oscillator is: \( x(t) = A e^{-\gamma t} \cos(\omega t + \phi) \) We apply the initial conditions, \(x(0) = 0 \) and \(\dot{x}(0) = v\), and find that: \( A = -\frac{v}{\omega} \) and \( \phi = \frac{\pi}{2} \) Our particular solution becomes: \( x(t) = -\frac{v}{\omega} \mathrm{e}^{-\gamma t} \cos(\omega t + \frac{\pi}{2}) \) Finally, we analyze the solution for a critically damped condition and find that as \(\omega_{0}\) approaches \(\gamma\), the solution tends to the corresponding solution for the critically damped oscillator: \( \lim_{\omega \to 0} x(t) \approx -\frac{v}{\omega}\sin(\omega t) \)

Step by step solution

01

Write down the oscillator equation

Begin by writing down the general equation of a damped harmonic oscillator: \(\ddot{x} + 2\gamma\dot{x} + \omega_{0}^2x = 0\) Given the initial conditions \(x(0) = 0\) and \(\dot{x}(0) = v\), we can solve this equation using the general solution for an underdamped oscillator, since \(\omega_{0} > \gamma\).
02

Write down the general solution for the underdamped oscillator

The general solution of an underdamped harmonic oscillator is given by: \(x(t) = A \mathrm{e}^{-\gamma t} \cos(\omega t + \phi)\) Where \(\omega = \sqrt{\omega_{0}^2 - \gamma^2}\), A represents the amplitude, and \(\phi\) is the phase angle. We need now to apply the initial conditions to find A and \(\phi\).
03

Apply the initial conditions to find A and \(\phi\)

Applying the initial conditions, we have: \(x(0) = A \mathrm{e}^0 \cos(0 + \phi) = A \cos\phi = 0\) Since \(\cos\phi = 0\), this implies that \(\phi = \frac{\pi}{2}\) (any odd multiple of \(\frac{\pi}{2}\) would also work, but we'll choose the simplest): \(\dot{x}(t) = -A\gamma \mathrm{e}^{-\gamma t} \cos(\omega t + \phi) - A\omega \mathrm{e}^{-\gamma t} \sin(\omega t + \phi)\) Applying the initial condition for the velocity: \(\dot{x}(0) = -A\gamma \cos\phi - A\omega \sin\phi = -A\omega = v\) \(A = -\frac{v}{\omega}\) Now, we have the particular solution for the given initial conditions: \(x(t) = -\frac{v}{\omega} \mathrm{e}^{-\gamma t} \cos(\omega t + \frac{\pi}{2})\)
04

Analyze the solution for critically damped condition

Now, we want to show that as \(\omega_{0}\) approaches the critical value \(\gamma\), the solution becomes the one for a critically damped oscillator. In the critically damped case: \(\omega_{0} = \gamma\) \(\omega = \sqrt{\gamma^2 - \gamma^2} = 0\) As \(\omega\) goes to zero, \(\frac{v}{\omega}\) can become very large, making the cosine term dominate the solution. For large values of A, we can write the solution as: \(x(t) \approx -\frac{v}{\omega}\cos(\omega t + \frac{\pi}{2})\) Let's take the limit as \(\omega\) approaches zero: \(\lim_{\omega \to 0} x(t) \approx \lim_{\omega \to 0} -\frac{v}{\omega}\cos(\omega t + \frac{\pi}{2})\) In the limit, the cosine term becomes: \(\lim_{\omega \to 0} \cos(\omega t + \frac{\pi}{2}) = \sin(\omega t)\) Substituting back into the expression for \(x(t)\) we get: \(\lim_{\omega \to 0} x(t) \approx -\frac{v}{\omega}\sin(\omega t)\) This expression is the same as the one obtained for a critically damped oscillator. Therefore, as \(\omega_{0}\) approaches \(\gamma\), the solution tends to the corresponding solution for the critically damped oscillator, as required.

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Key Concepts

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

Underdamped Oscillator Solution
Understanding the behavior of a damped harmonic oscillator is pivotal in physics, especially when dealing with systems that oscillate and gradually come to a rest due to resistive forces, like friction or air resistance. An underdamped oscillator solution is particularly important as it represents a system in which the damping is not strong enough to prevent oscillations.

In mathematical terms, when the natural frequency of the system \(\omega_{0}\) is greater than the damping coefficient \(\gamma\), the system is underdamped and the solution to the oscillator equation reflects this by showing a combination of decaying exponential behavior and sinusoidal oscillation. The general solution for such a system can be expressed as:
\[x(t) = A e^{-\gamma t} \cos(\omega t + \phi)\]
Here, \(A\) represents the amplitude of the oscillation, \(\omega\) is the damped natural frequency, and \(\phi\) is the phase constant. The damped natural frequency is calculated as \(\omega = \sqrt{\omega_{0}^{2} - \gamma^{2}}\).

The inclusion of the exponential term \(e^{-\gamma t}\) reflects how the amplitude of the oscillations diminishes over time due to the damping forces at play. Students often find visualizing the graph of this function helpful, as it clearly shows oscillations that progressively decrease in amplitude. In practical applications, this type of motion might be observed in systems like car suspensions or the strings of musical instruments.
Critical Damping
Critical damping occurs at the precise point where the damping force is strong enough to prevent oscillations but not too strong to over-dampen the system, resulting in the quickest return to equilibrium without oscillating. This concept is integral in systems design, where engineers strive for rapid stabilization without overshoot.

Mathematically, a system is critically damped when the damping coefficient \(\gamma\) equals the natural frequency \(\omega_{0}\). At this point, the solution to the displacement \(x(t)\) of a critically damped oscillator is not oscillatory, unlike the underdamped case. Instead, it is a smooth decay back to equilibrium, typified by a rapid decrease in amplitude.

An interesting aspect of critical damping is that as \(\omega_{0}\) approaches \(\gamma\) in an underdamped oscillator, the sinusoidal nature of the solution diminishes and the characteristic equation of the system no longer has complex roots, leading to a solution involving real exponentials. The solution of a critically damped oscillator can be expressed as:
\[x(t) = (C_1 + C_2t)e^{-\gamma t}\]
With the correct initial conditions, the system will move back to the equilibrium position as rapidly as possible without overshooting, characteristic of critical damping.
Oscillator Initial Conditions
The initial conditions of a harmonic oscillator define its state at the beginning of the motion and are crucial for determining its subsequent behavior. These are typically given in terms of initial displacement \(x(0)\) and initial velocity \(\dot{x}(0)\). In the context of the damped harmonic oscillator, applying these initial conditions allows us to find specific solutions to the equation of motion that satisfy the given starting state of the system.

For example, if an underdamped oscillator starts at rest from a certain displacement, the initial conditions would be \(x(0) = x_{0}\) and \(\dot{x}(0) = 0\). On the other hand, if it starts from the equilibrium position with an initial velocity, the initial conditions would be \(x(0) = 0\) and \(\dot{x}(0) = v\), as seen in the given exercise. By applying these specific initial conditions to the general solution, one can derive coefficients that tailor the generic solution to the particular scenario of the system's initial state.

It is the application of these conditions that make solving physics problems practical; they allow us to predict the motion of systems under various starting scenarios. Teaching students to correctly apply these initial conditions is therefore fundamental in fostering a deeper understanding of physical systems and their analysis.

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Most popular questions from this chapter

*The potential energy function of a particle of mass \(m\) is \(V=c x /\left(x^{2}+\right.\) \(\left.a^{2}\right)\), where \(c\) and \(a\) are positive constants. Sketch \(V\) as a function of \(x\). Find the position of stable equilibrium, and the period of small oscillations about it. Given that the particle starts from this point with velocity \(v\), find the ranges of values of \(v\) for which it (a) oscillates, (b) escapes to \(-\infty\), and \((c)\) escapes to \(+\infty\).

Solve the problem of an oscillator under a simple periodic force (turned on at \(t=0\) ) by the Green's function method, and verify that it reproduces the solution of \(\S 2.6\). [Assume that the damping is less than critical. To do the integral, write \(\sin \omega t\) as \(\left(\mathrm{e}^{\mathrm{i} \omega t}-\mathrm{e}^{-\mathrm{i} \omega t}\right) / 2 \mathrm{i}\).]

A particle moves vertically under gravity and a retarding force proportional to the square of its velocity. (This is more appropriate than a linear relation for larger particles - see Problem 8.10.) If \(v\) is its upward or downward speed, show that \(\dot{v}=\mp g-k v^{2}\), respectively, where \(k\) is a constant. If the particle is moving upwards, show that its position at time \(t\) is given by \(z=z_{0}+(1 / k) \ln \cos \left[\sqrt{g k}\left(t_{0}-t\right)\right]\), where \(z_{0}\) and \(t_{0}\) are integration constants. If its initial velocity at \(t=0\) is \(u\), find the time at which it comes instantaneously to rest, and its height then. [Note that \(\ln\) always denotes the natural logarithm: \(\ln x \equiv \log _{\mathrm{e}} x\). You may find the identity \(\ln \cos x=-\frac{1}{2} \ln \left(1+\tan ^{2} x\right)\) useful.]

The potential energy function of a particle of mass \(m\) is \(V=-\frac{1}{2} c\left(x^{2}-\right.\) \(\left.a^{2}\right)^{2}\), where \(c\) and \(a\) are positive constants. Sketch this function, and describe the possible types of motion in the three cases (a) \(E>0\), (b) \(E<-\frac{1}{2} c a^{4}\), and \((\mathrm{c})-\frac{1}{2} c a^{4}

A particle moving under a conservative force oscillates between \(x_{1}\) and \(x_{2}\). Show that the period of oscillation is $$ \tau=2 \int_{x_{1}}^{x_{2}} \sqrt{\frac{m}{2\left[V\left(x_{2}\right)-V(x)\right]}} \mathrm{d} x $$ In particular, if \(V=\frac{1}{2} m \omega_{0}^{2}\left(x^{2}-b x^{4}\right)\), show that the period for oscillations of amplitude \(a\) is $$ \tau=\frac{2}{\omega_{0}} \int_{-a}^{a} \frac{\mathrm{d} x}{\sqrt{a^{2}-x^{2}} \sqrt{1-b\left(a^{2}+x^{2}\right)}} $$ Using the binomial theorem to expand in powers of \(b\), and the substitution \(x=a \sin \theta\), show that for small amplitude the period is approximately \(\tau \approx 2 \pi\left(1+\frac{3}{4} b a^{2}\right) / \omega_{0}\).

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