Question

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum.) Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3}\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this ascillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x| / \alpha\), where \(\alpha\) is any positive value and \(\gamma\) is a constant

Short answer

The equation of motion for the oscillator is given by: $$\frac{d^2x}{dt^2} = -\frac{c}{m}x^3.$$ And the period of the oscillation depends on the amplitude \(A\) as: $$T \propto A^{\alpha},$$ where \(\alpha\) can be determined by examining the integral expression for the period.

Step-by-step solution

Derive the equation of motion

The equation of motion for the oscillator can be written using Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object. In this case, the net force is given by the restoring force, \(F(x)=-c x^{3}\): $$m \frac{d^2x}{dt^2} = -cx^3,$$ so we have $$\frac{d^2x}{dt^2} = -\frac{c}{m}x^3.$$

Find the period of the oscillator

To find the period of the oscillation, we can first integrate the equation of motion to find the time it takes the oscillator to travel from one extreme point to the other and then multiply it by 2. When the oscillator reaches the extreme points, the velocity is zero. We have: \begin{align*} v = \frac{dx}{dt} = \pm \sqrt{A^2 - x^2}, \end{align*} where \(A\) is the amplitude of oscillation. Next, we need to find the time as the extreme point: $$t=\int_0^A \frac{dx}{\sqrt{A^2-x^2}}.$$ Now, we can find the period by multiplying the time by 2: $$T = 2 \int_0^A \frac{dx}{\sqrt{A^2-x^2}}.$$

Determine the dependence of the period on the amplitude

To find the relationship between the amplitude and the period, we can rewrite the period integral as: $$T =\frac{ 2}{\sqrt{c/m}} \int_0^A \frac{dx}{\sqrt{(A^2-x^2)^{3/2}}}.$$ Now we can see that the period depends on the amplitude A, and the relationship is given by $$T \propto A^{\alpha},$$ where \(\alpha\) can be determined by looking at the expression inside the integral.

Generalize the results

Now we consider a potential energy given by \(U(x)=\gamma|x| / \alpha\). The restoring force will be the negative gradient of the potential energy: $$F(x) = - \frac{dU}{dx} = -\gamma \frac{d|x|^{\alpha}}{dx}.$$ We can now repeat the steps 1 through 3 using this new potential energy and force, and determine the period and the relationship between the amplitude and the period. In summary, we found the period integral expression for an oscillator with a restoring force given by \(F(x)=-c x^{3}\), and derived the relation between the period and the amplitude. We then generalized the result for a potential energy given by \(U(x)=\gamma|x| / \alpha\).

Key concepts

Oscillator Period

In the realm of physics, particularly when studying oscillatory motion, the term oscillator period plays a central role. It's defined as the time it takes for one complete cycle of motion to occur. Imagine a swing swinging back and forth; the period is the duration from one peak to back to that peak again. When we consider simple harmonic motion, a common example of an oscillator, Newton's second law is often used to derive the oscillator's equation of motion.

For a classic simple harmonic oscillator where the force is directly proportional to the displacement, the period is famously independent of the amplitude of the oscillation. This principle, known as isochrony, suggests that no matter how far you pull the swing back, it will always take the same time to complete a cycle, assuming no external forces like friction are at play. However, for more complex forces such as the one given in the exercise, where the restoring force is proportional to the cube of the displacement, the period does indeed become amplitude-dependent, deviating from isochrony. The period of such an oscillator is found by integrating the equation of motion, specifically by calculating the time for the object to go from one extreme of its motion to the other and then doubling it.

Restoring Force

The concept of a restoring force is an essential one in the study of oscillations. As the name suggests, this is the force that acts to restore an object to its equilibrium position. Imagine stretching a rubber band; the more you stretch, the stronger the force pulling it back to its original shape. In simple harmonic motion, the restoring force is typically linear, which means it's directly proportional to the displacement from the equilibrium position---think of it as the rubber band's consistent pull.

In our exercise, the restoring force is a bit more unusual, defined as \( F(x)=-c x^{3} \). Unlike the simple linear case, this force increases more dramatically as the object moves further from equilibrium, since it's dependent on the cube of the displacement. This non-linearity leads to complexities in the motion of the oscillator. The equation of motion integration step that we undertake next helps to predict how the oscillator will move in response to such a force.

Oscillation Amplitude-Dependence

With simple harmonic oscillators, it's often noted that the amplitude of oscillation does not affect the period. This property might not hold for other types of oscillatory motion, where the oscillation amplitude-dependence becomes a key factor. The amplitude refers to the maximum extent of the vibration or the furthest distance from the equilibrium position.

In the given exercise, the oscillator's period is indeed amplitude-dependent, evidenced by the equation \( T = 2 \int_0^A \frac{dx}{\sqrt{A^2-x^2}} \), where the integral's limit includes the amplitude \( A \). By examining the integral, we can see that the period's dependence on the amplitude isn't straightforward and will require a deeper analysis. Interestingly, this relationship also implies that larger amplitudes would lead to longer periods, which is a departure from the isochrony exhibited by classic simple harmonic motion.

Equation of Motion Integration

To fully understand the motion of oscillators, one often needs to perform an equation of motion integration. This process involves taking the equation that describes the motion of the oscillator and integrating it to find displacement, velocity, or time over one or more cycles of motion. Integration, a fundamental tool in calculus, provides the means to calculate quantities that are variable over time, such as the area under a curve.

In our exercise, integration allows us to find the time taken for the oscillator to move between two extreme points of its motion. The integration of \( \frac{dx}{\sqrt{A^2-x^2}} \) is tied to the period of the oscillator and is used to extrapolate how changes in the amplitude affect the motion. Determining the period from this integral for non-linear restoring forces, unlike linear forces, requires careful assessment of the integral's form and limits. The complexity of equation integration in these scenarios is paramount to gaining a complete picture of the oscillator's behavior.