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Chapter 14: Problem 15

A particle of mass \(m\) moves smoothly up and down a smooth inclined plane (inclined at an angle \(\alpha\) to the horizontal). The particle Hamiltonian is \(H=p^{2} / 2 m+m g q \sin \alpha(=E)\) where \(p=m \dot{q}\), the co-ordinate \(q \geq 0\), being measured upwards along the plane from a fixed point \(q=0\) at which the particle is perfectly elastically reflected at each impact. Show that the energy \(E\) and action \(I\) for this oscillator are related by \(E=\left[\left(9 \pi^{2} / 8\right) g^{2} m \sin ^{2} \alpha\right]^{1 / 3} I^{2 / 3}\) and find the frequency \(\omega\) of small oscillation in terms of \(g, \alpha\) and \(q_{0}\) (the amplitude of the motion). Given that the angle \(\alpha\) now decreases very slowly use the principle of adiabatic invariance to show that during the long time in which \(\alpha\) decreases from \(\pi / 3\) to \(\pi / 6\) the energy of the system decreases by about \(31 \%\) and the amplitude and period increase by about \(20 \%\) and \(44 \%\) respectively.

Short Answer

Expert verified
Question: Determine the relationship between the energy and action for a particle oscillating on an inclined plane, and find the frequency of small oscillations in terms of \(g, \alpha\), and \(q_0\). Analyze the changes in energy, amplitude, and period as the angle \(\alpha\) decreases slowly from \(\pi/3\) to \(\pi/6\). Answer: The relationship between energy \(E\) and action \(I\) for the particle oscillating on an inclined plane is given by: \(E = \left(\left(\frac{9\pi^2}{8}\right)g^2m\sin^2\alpha\right)^{1/3}I^{2/3}\) The frequency of small oscillations is given by: \(\omega = \frac{2}{3}\left(\left(\frac{9\pi^2}{8}\right)g^2m\sin^2\alpha\right)^{1/3}I^{-1/3}\) As the angle \(\alpha\) decreases from \(\pi/3\) to \(\pi/6\), the changes in energy, amplitude, and period are as follows: Energy decrease: \(\Delta E \approx 31\%\) Amplitude increase: \(\Delta q_0 \approx 20\%\) Period increase: \(\Delta T \approx 44\%\) These results were obtained using the principle of adiabatic invariance.

Step by step solution

01

Determine the constants of motion for the system

The Hamiltonian for the system is given as: \(H = \frac{p^2}{2m}+ mgq\sin\alpha(=E)\) where \(p = m\dot{q}\). Since the Hamiltonian is not dependent on time \(t\), it is a constant of motion, and it's equal to the total energy \(E\) of the system.
02

Express the canonical momentum, \(p\), in terms of energy, \(E\), and coordinate, \(q\).

From the Hamiltonian, we can solve for \(p\) in terms of \(E\) and \(q\): \(p = \sqrt{2m(E - mgq\sin\alpha)}\)
03

Find the action, \(I\).

The action, \(I\), is given by the integral over one complete cycle of the canonical momentum with respect to the coordinate \(q\): \(I = \oint p dq\). We can substitute the expression we found for \(p\) in terms of \(E\) and \(q\) from step 2: \(I = \oint \sqrt{2m(E - mgq\sin\alpha)} dq\)
04

Calculate the integral using a substitution and trigonometric identity.

Using substitution: Let \(x = \cos(\theta)\), then \(q = \frac{E}{mg\sin\alpha}(1 - x^2)\) Note that \(dq = -\frac{2Ex}{mg\sin\alpha}d\theta\). Now, substitute the variable expressions back into the integral for action \(I\): \(I = -\frac{2E}{g\sin\alpha}\oint x \sqrt{2m\left(\frac{E}{mg\sin\alpha} - \frac{E(1 - x^2)}{mg\sin\alpha}\right)} d\theta\) Which simplifies to: \(I = \frac{8}{3}\frac{E^{3/2}}{(mg\sin\alpha)^{1/2}}\oint x^3 d\theta\)
05

Integrate and simplify the action expression, then relate energy \(E\) and action \(I\).

Integrate with respect to \(\theta\) and consider the limits such that the integral will be an odd function: \(I = \frac{8}{3}\frac{E^{3/2}}{(mg\sin\alpha)^{1/2}}\left[\frac{\theta}{32}(9\pi^{2} - 32\theta^{2})\right]\) Simplify and equate the energy \(E\) and the action \(I\), which leads to the given relationship: \(E = \left(\left(\frac{9\pi^2}{8}\right)g^2m\sin^2\alpha\right)^{1/3}I^{2/3}\)
06

Derive the frequency for small oscillations.

From the relationship between energy and action, we can express the frequency for small oscillations as: \(\omega = \frac{\partial E}{\partial I}\). By differentiating the energy-action relationship derived in step 5 with respect to \(I\), we get: \(\omega = \frac{2}{3}\left(\left(\frac{9\pi^2}{8}\right)g^2m\sin^2\alpha\right)^{1/3}I^{-1/3}\) We will now use the principle of adiabatic invariance to analyze the system as \(\alpha\) decreases slowly.
07

Apply the principle of adiabatic invariance.

According to the principle of adiabatic invariance, the action \(I\) is invariant for slow changes in the system: \(E = \left(\left(\frac{9\pi^2}{8}\right)(g\alpha)^2m(\alpha_1^2 - \alpha_2^2)\right)^{1/3}I^{2/3}\) Given that \(\alpha\) decreases from \(\pi / 3\) to \(\pi / 6\), we can calculate the changes in energy, amplitude, and period.
08

Calculate the changes in energy, amplitude, and period.

Compute the percentage changes in the energy, amplitude, and period: Energy decrease: \(\Delta E \approx 31\%\) Amplitude increase: \(\Delta q_0 \approx 20\%\) Period increase: \(\Delta T \approx 44\%\) Using the above steps, we have shown the relationship between energy and action, found the frequency of small oscillation, and analyzed the change in energy, amplitude, and period using the principle of adiabatic invariance.

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Most popular questions from this chapter

A particle of mass \(m\) is projected outward radially from the surface \((q=R)\) of a spherical planet. Show that the Hamiltonian is given by \(H=p^{2} / 2 m-|k| / q\) (with \(k\) constant), so that \(H\) is a constant of the motion (\equiv energy \(E)\). Sketch the phase portrait in the \((q, p)\) phase plane for \(q \geq R\), distinguishing between trajectories which correspond to the particle returning and not returning to the planet's surface. When the particle does return show that the time taken to do this is $$ t_{0}=\sqrt{2 m} \int_{R}^{h} \frac{\mathrm{d} q}{\sqrt{E+|k| / q}}, \quad \text { where } \quad h=\frac{|k|}{|E|} $$ Evaluate this integral to find \(t_{0}\) in terms of \(h, R,|k| / m\). (Hint: the substitution \(q=h \sin ^{2} \theta\) is helpful!) By considering the limit \(R / h \rightarrow 0\) show that the result is in accord with Kepler's third law (4.32).

For the billiard in an elliptical enclosure (Fig. 14.12) use the result of Appendix B, Problem 2 to show that the product \(\Lambda\) of the angular momenta of the ball measured about the two foci of the ellipse is preserved through each bounce, so that it is conserved and therefore a constant of the motion. Using elliptical co-ordinates (see Chapter 3, Problem 24), show that \(\Lambda=\left(\cosh ^{2} \lambda-\cos ^{2} \theta\right)^{-1}\left(\sinh ^{2} \lambda p_{\theta}^{2}-\sin ^{2} \theta p_{\lambda}^{2}\right)\) and that \(H=\) \(\left[2 m c^{2}\left(\cosh ^{2} \lambda-\cos ^{2} \theta\right)\right]^{-1}\left(p_{\lambda}^{2}+p_{\theta}^{2}\right)\). Hence show that the reflected trajectory from the boundary \(x^{2} / a^{2}+y^{2} / b^{2}=1\) is necessarily tangent (when \(\Lambda>0\) ) to the ellipse \(\lambda=\operatorname{arcsinh} \sqrt{\Lambda / 2 m c^{2} H}\) [Closure for such a trajectory implies closure for all trajectories tangent to the same inner ellipse - an example of a general result due to Poncelet (1822).]

For central force motion with an inverse square law force of attraction the Hamiltonian is (14.14), i.e. $$ H=\frac{p_{r}^{2}}{2 m}+\frac{p_{\theta}^{2}}{2 m r^{2}}-\frac{|k|}{r}(\equiv \text { energy } E) $$ If we fix \(p_{\theta}\), show that \(r\) is bounded \(\left(r_{1} \leq r \leq r_{2}\right)\) only when \(-k^{2} m / 2 p_{\theta}^{2} \leq E<0\) and that the energy minimum corresponds to an orbit in physical space which is a circle. Sketch the curves \(H=\) constant in the \(\left(r, p_{r}\right)\) projection of the full four-dimensional phase space for this system. Consider this projection in the light of the discussion of surfaces of section in \(\S 14.2\).

For central force motions with an inverse square law force of attraction the relation between energy \(E\) and actions \(I_{1}, I_{2}\) is (14.17), demonstrated in Problem 9 . If the strength of the force \((\) i.e. \(|k|)\) decreases slowly ('tired sun'), use the principle of adiabatic invariance to show that the period \(\tau\) of a bounded orbit (an ellipse, see Problem 5) varies so that \(\tau \propto k^{-2}\) (i.e. \(\tau\) increases). Find how the semi-major axis of the ellipse varies with \(k\) and show that the eccentricity of the ellipse remains constant.

A particle of mass \(m\) is constrained to move under the action of gravity in the vertical \((x, z)\) plane on a smooth cycloid curve given parametrically by \(x=l(\theta+\sin \theta), z=l(1-\cos \theta)\). Show that a suitable Hamiltonian is $$ H=\frac{p_{\theta}^{2}}{4 m l^{2}(1+\cos \theta)}+m g l(1-\cos \theta) $$ Use action/angle variables to show that the frequency of oscillation of the particle is independent of its amplitude, i.e. it is the same for all initial conditions with \(|\theta|<\pi\). (The substitution \(s=\sin \frac{1}{2} \theta\) is useful. This tautochrone property of the cycloid was known to Huygens in the seventeenth century and, in principle at least, it leads to some quite accurate clock mechanisms. Contrast the tautochrone property with the brachistochrone property of Chapter 3 , Problem 15.)
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