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In order to prove the crucial formulas (5.83)?5.85) for the Fourier coefficients \(a_{n}\) and \(b_{n},\) you must first prove the following: $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \cos (m \omega t) d t=\left\\{\begin{array}{ll} \tau / 2 & \text { if } m=n \neq 0 \\ 0 & \text { if } m \neq n \end{array}\right.$$ (This integral is obviously \(\tau\) if \(m=n=0 .\) ) There is an identical result with all cosines replaced by sines, and finally $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \sin (m \omega t) d t=0 \quad \text { for all integers } n \text { and } m$$ where as usual \(\omega=2 \pi / \tau\). Prove these. [Hint: Use trig identities to replace \(\cos (\theta) \cos (\phi)\) by terms like \(\cos (\theta+\phi) \text { and so on. }]\)

Short Answer

Expert verified
Use trig identities to show integrals yield specific values or zero based on \(m\) and \(n\) values.

Step by step solution

01

Understand the Function to Integrate

We need to integrate functions of the form \(\cos(n \omega t) \cos(m \omega t)\) over the interval \([-\tau/2, \tau/2]\). The result depends on whether \(n\) is equal to \(m\) or not, and whether \(n\) and \(m\) are both zero or not.
02

Apply Trigonometric Identity

Use the identity \(\cos(A) \cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\) to simplify the integral. Replace \( A = n\omega t \) and \( B = m\omega t \) to get \[\cos(n \omega t) \cos(m \omega t) = \frac{1}{2}[\cos((n+m)\omega t) + \cos((n-m)\omega t)].\]
03

Integrate the Resulting Expressions

Split the integral into two separate parts, each involving the cosine of a single angle: \[ \int_{-\tau/2}^{\tau/2} \cos(n \omega t) \cos(m \omega t) \, dt = \frac{1}{2} \int_{-\tau/2}^{\tau/2} \cos((n+m)\omega t) \, dt + \frac{1}{2} \int_{-\tau/2}^{\tau/2} \cos((n-m)\omega t) \, dt. \] Evaluate each of these integrals.
04

Evaluate Special Cases

Consider the three cases: (i) \(n = m = 0\), (ii) \(m = n eq 0\), and (iii) \(m eq n\).- For case (i): \(\cos(0) = 1\), the integral becomes \(\int_{-\tau/2}^{\tau/2} 1 \, dt = \tau.\)- For case (ii): Both terms simplify to \(\int_{-\tau/2}^{\tau/2} \cos(0) \, dt = \frac{\tau}{2}.\)- For case (iii): Both integrals evaluate to zero due to the periodic nature of cosines with frequencies not equal to zero.
05

Verify Results with Sines

Perform similar steps replacing cosines with sines. Use the identity \( \sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \) to simplify, and verify that the integral results in 0 for \(\int_{-\tau/2}^{\tau/2} \sin(n \omega t) \sin(m \omega t) \, dt \), again resulting in zero unless both are zero, which does not apply here.
06

Integrate Cosine and Sine Product

Lastly, consider \( \int_{-\tau/2}^{\tau/2} \cos(n \omega t) \sin(m \omega t) \, dt \). Since \( \cos \) and \( \sin \) are orthogonal over a full period, any such integral is zero. This works for all integer \( n \) and \( m \), confirming that the result is zero.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are the building blocks for simplifying expressions involving trigonometric functions. They provide relationships between functions like sine, cosine, and tangent that allow us to transform complex expressions into simpler ones. In our case, we used the identity:
  • \(\cos(A) \cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\)
This identity lets us express the product of two cosine functions as a sum of two other cosines.
This simplification is particularly useful when integrating products of trigonometric functions. By breaking down the original product into simpler terms, we can more easily compute integrals, which form the core of finding Fourier coefficients.
It's crucial to understand and be able to derive these identities, as they enable us to solve complex mathematical problems more efficiently.
Orthogonality
Orthogonality in the context of trigonometric functions refers to certain pairs of functions being mutually "independent" over specific intervals. This concept is central in Fourier analysis.
Essentially, for functions defined over the same interval, the integral of their product can result in zero, which mathematically states their independence or orthogonality. Key points are:
  • \(\int_{-\tau/2}^{\tau/2} \cos(n \omega t) \sin(m \omega t) \, dt = 0\) for all integers \(n\) and \(m\).
  • \(\int_{-\tau/2}^{\tau/2} \cos(n \omega t) \cos(m \omega t) \, dt = 0\) if \(m eq n\).
This principle of orthogonality is crucial when dealing with series like Fourier series, as it simplifies the integration process by canceling out terms that would typically complicate calculations.
Understanding orthogonality helps clarify why certain terms vanish in integrations, thus simplifying the solution of differential equations and the decomposition of signals into their frequency components.
Integrals
Integrals are fundamental in calculus and are used to find quantities like area, volume, and in this context, Fourier coefficients. The concept involves accumulating quantities over a range, and the integral of a function over a specific interval gives an aggregate measure of that function along that interval. In our problem:
  • \(\int_{-\tau/2}^{\tau/2} 1 \, dt = \tau\), which finds the 'area' under a flat line across the interval.
  • \(\int_{-\tau/2}^{\tau/2} \cos((n+m)\omega t) \, dt\) and \(\int_{-\tau/2}^{\tau/2} \cos((n-m)\omega t) \, dt\) simplify due to properties of periodic functions, often resulting in zero unless special conditions are met.
Understanding how to evaluate these integrals allows us to compute the Fourier coefficients that represent periodic functions in terms of sine and cosine functions.
This understanding is vital for signal processing, systems analysis, and solving physics problems where periodic functions play a pivotal role.

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Most popular questions from this chapter

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?

Consider a cart on a spring with natural frequency \(\omega_{\mathrm{o}}=2 \pi,\) which is released from rest at \(x_{\mathrm{o}}=1\) and \(t=0 .\) Using appropriate graphing software, plot the position \(x(t)\) for \(0 < t < 2\) and for damping constants \(\beta=0,1,2,4,6,2 \pi, 10,\) and \(20 .\) [Remember that \(x(t)\) is given by different formulas for \(\left.\beta<\omega_{\mathrm{o}}, \beta=\omega_{\mathrm{o}}, \text { and } \beta > \omega_{\mathrm{o}} .\right]\)

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

Another interpretation of the \(Q\) of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set \(\omega=\omega_{\mathrm{o}}\). (a) Show that the oscillator's total energy (kinetic plus potential) is \(E=\frac{1}{2} m \omega^{2} A^{2} .\) (b) Show that the energy \(\Delta E_{\text {dis dissipated during one cycle by the damping force }}\) \(F_{\text {dmp }}\) is \(2 \pi m \beta \omega A^{2} .\) (Remember that the rate at which a force does work is \(F v .\) ) (c) Hence show that \(Q\) is \(2 \pi\) times the ratio \(E / \Delta E_{\text {dis: }}\)

(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?

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