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Let \(f(t)\) be a periodic function with period \(\tau\). Explain clearly why the average of \(f\) over one period is not necessarily the same as the average over some other time interval. Explain why, on the other hand, the average over a long time \(T\) approaches the average over one period, as \(T \rightarrow \infty\)

Short Answer

Expert verified
The average over one period may differ from other intervals, but averages converge as time interval grows longer.

Step by step solution

01

Understanding a Periodic Function

A periodic function, like \(f(t)\), repeats its values in regular intervals, known as its period \(\tau\). This means that the behavior of \(f(t)\) is consistent over each segment of length \(\tau\), making \(f(t + \tau) = f(t)\) for any \(t\).
02

Average Over One Period

The average value of \(f(t)\) over one period \(\tau\) is calculated as \(\frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt\). This represents how the values of \(f(t)\) are distributed over a single cycle.
03

Difference with Average Over Other Intervals

If we take an average over a time interval different from \(\tau\), say from \(0\) to \(T\) (where \(T\) is not a multiple of \(\tau\)), the function \(f(t)\) may not complete its periods entirely within \(T\). Thus, some parts of \(f(t)\) might be overrepresented or underrepresented, leading to an average that differs from the one-period average.
04

Approaching the Period Average as Time Increases

When taking the average over a long time \(T\), the interval \(T\) can potentially include many full periods of \(f(t)\). As \(T \rightarrow \infty\), any discrepancies from incomplete periods at the beginning or end of the interval become negligible. Thus, the average value over \(T\) approaches the average over a complete period. This can be expressed as \(\frac{1}{T}\int_{0}^{T} f(t) \, dt \approx \frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt \) as \(T \rightarrow \infty\).

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Key Concepts

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

Average Value Over a Period
In mathematics, the average value of a periodic function over one complete period is a foundational concept. When we talk about periodic functions like \( f(t) \), which consistently repeat their behavior every period \( \tau \), this periodicity ensures predictability over each cycle. The average is obtained by integrating the function over one entire period and then dividing by the period length, given by the formula: \[ \text{Average over one period} = \frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt \]This calculation tells us how the function behaves over each individual cycle. It assumes that each possible value within a cycle contributes proportionally to the average. For instance, if \( f(t) \) represents a sine wave, the average over one full sine wave cycle would consider both the ups and downs balanced over a complete period.
Incomplete Periods
When a periodic function is averaged over an interval that does not line up perfectly with its period, such as \( T \) where \( T eq n\tau \) for any integer \( n \), we encounter the challenge of incomplete periods. Imagine slicing through a loaf of bread—the slices don’t complete a full loaf exactly, leading to uneven pieces. Similarly, in an incomplete period, parts of \( f(t) \) might not be represented fully. This results in some values being counted more than others, while others may be missed completely, skewing the average. For example, if our interval stops halfway through an oscillation, the average becomes biased towards that partially completed oscillation. This difference occurs because the function might not return to its starting point, creating an imbalance in representation.
Long-Term Averaging
Long-term averaging is like observing a constant drip turning into a steady stream. When you average \( f(t) \) over a very large interval \( T \), the partial periods at start and end become increasingly insignificant. This is because the ratio of partial periods to complete periods decreases as \( T \) becomes very large. Thus, expressed mathematically, the long-term average:\[ \lim_{T \to \infty} \frac{1}{T} \int_{0}^{T} f(t) \, dt = \frac{1}{\tau} \int_{0}^{\tau} f(t) \, dt \]ensures the function’s true periodic nature eventually dominates any initial irregularities. This situation reflects the principle of ergodicity, where time averages and ensemble averages yield the same result in the limit, assuming the function is reasonably well-behaved. By considering ever longer times, you "smooth out" any discrepancies caused by incomplete cycles, converging towards the true average of the period.

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Most popular questions from this chapter

(a) Consider a cart on a spring which is critically damped. At time \(t=0\), it is sitting at its equilibrium position and is kicked in the positive direction with velocity \(v_{\mathrm{o}} .\) Find its position \(x(t)\) for all subsequent times and sketch your answer. (b) Do the same for the case that it is released from rest at position \(x=x_{\mathrm{o}} .\) In this latter case, how far is the cart from equilibrium after a time equal to \(\tau_{\mathrm{o}}=2 \pi / \omega_{\mathrm{o}}\) the period in the absence of any damping?

An unusual pendulum is made by fixing a string to a horizontal cylinder of radius \(R\), wrapping the string several times around the cylinder, and then tying a mass \(m\) to the loose end. In equilibrium the mass hangs a distance \(l_{\mathrm{o}}\) vertically below the edge of the cylinder. Find the potential energy if the pendulum has swung to an angle \(\phi\) from the vertical. Show that for small angles, it can be written in the Hooke's law form \(U=\frac{1}{2} k \phi^{2} .\) Comment on the value of \(k\).

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 \(\left.\omega+\omega_{\mathrm{o}} \approx 2 \omega_{\mathrm{o}}, \text { but you certainly mustn't say } \omega-\omega_{\mathrm{o}} \approx 0 .\right]\)

The maximum displacement of a mass oscillating about its equilibrium position is \(0.2 \mathrm{m},\) and its maximum speed is \(1.2 \mathrm{m} / \mathrm{s}\). What is the period \(\tau\) of its oscillations?

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four \(80-\mathrm{kg}\) men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant \(k\) of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass \(50 \mathrm{kg}\), what is the natural frequency of the assembly oscillating on its two springs? ( \(\mathbf{c}\) ) If the bumps on a road are \(80 \mathrm{cm}\) apart, at about what speed would these oscillations go into resonance?

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