/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 (a) Use complex number technique... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Use complex number techniques to rewrite the function \(f(t)=4 \sin \omega t+3 \cos \omega t\) in the form \(A \sin (\omega t+\delta)\).(answer check available at lightandmatter.com) (b) Verify the result using the trigonometric identity \(\sin (\alpha+\beta)=\sin \alpha \cos \beta+\sin \beta \cos \alpha\).

Short Answer

Expert verified
The function is rewritten as \(5 \sin(\omega t + \tan^{-1}(4/3))\).

Step by step solution

01

Express Sine and Cosine in Terms of Complex Exponentials

To rewrite the function, use the identities: \( \sin \omega t = \frac{e^{i \omega t} - e^{-i \omega t}}{2i} \) and \( \cos \omega t = \frac{e^{i \omega t} + e^{-i \omega t}}{2} \). Substitute these into \( f(t) = 4 \sin \omega t + 3 \cos \omega t \) to express each term as a complex exponential.
02

Combine and Simplify the Expression

Substituting these identities, we have: \( f(t) = 4 \left( \frac{e^{i \omega t} - e^{-i \omega t}}{2i} \right) + 3 \left( \frac{e^{i \omega t} + e^{-i \omega t}}{2} \right) \). Simplify to combine like terms, yielding: \( f(t) = \left( \frac{3}{2} - 2i \right) e^{i \omega t} + \left( \frac{3}{2} + 2i \right) e^{-i \omega t} \).
03

Determine Amplitude and Phase

Recognize that this can be expressed as a single sinusoidal function: \( f(t) = A \sin(\omega t + \delta) \). Calculate \( A \) using \( A = \sqrt{a^2 + b^2} \), where \( a = 3 \) and \( b = 4 \). Thus, \( A = 5 \). Use \( \tan \delta = \frac{b}{a} \) to find the phase angle \( \delta \), so \( \tan \delta = \frac{4}{3} \) with \( \delta = \tan^{-1}\left(\frac{4}{3}\right) \).
04

Verify Using Trigonometric Identity

Verify the result by applying the identity: \( \sin(\omega t + \delta) = \sin \omega t \cos \delta + \cos \omega t \sin \delta \). Calculating: \( f(t) = 5(\sin\omega t \cdot \frac{3}{5} + \cos \omega t \cdot \frac{4}{5}) = 3 \cos \omega t + 4 \sin \omega t \), matching the original function form.

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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 essential tools in simplifying and transforming expressions involving trigonometric functions. One of the most useful identities in this realm is the angle addition identity for sine:
  • \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
This identity allows us to express the sine of a sum of two angles in terms of products of sines and cosines. In the context of complex numbers and solving the problem provided, applying this identity helps verify transformed expressions.
For example, when dealing with a function like \( f(t) = A \sin(\omega t + \delta) \), breaking it down into \( \sin \omega t \cos \delta + \cos \omega t \sin \delta \) using the trigonometric identity can show the equivalence to expressions like \( 4 \sin \omega t + 3 \cos \omega t \), thus proving the manipulation correctness through known identities.
Complex Exponentials
Complex exponentials can be instrumental in simplifying trigonometric functions. Typically, we express sines and cosines utilizing the complex exponential form through Euler's formula:
  • \( e^{i \omega t} = \cos \omega t + i \sin \omega t \)
  • \( \sin \omega t = \frac{e^{i \omega t} - e^{-i \omega t}}{2i} \)
  • \( \cos \omega t = \frac{e^{i \omega t} + e^{-i \omega t}}{2} \)
By substituting these expressions into a trigonometric function, we can often convert a problem into a more manageable form using algebraic manipulation, which simplifies to arithmetic with complex numbers.
In our problem, representing \( 4 \sin \omega t + 3 \cos \omega t \) as a combination of complex exponentials leads to a simplified expression that can be expressed back as a single sinusoidal function, \( A \sin(\omega t + \delta) \). This process not only makes calculations simpler but also highlights underlying symmetries between trigonometric and exponential functions.
Amplitude and Phase
Amplitude and phase are key characteristics of sinusoidal waveforms, revealing information about a wave's strength and shift. The amplitude, \( A \), measures the peak value of the wave, while the phase, \( \delta \), indicates the horizontal shift along the time axis.
  • Amplitude calculation: \( A = \sqrt{a^2 + b^2} \)
  • Phase angle: \( \tan \delta = \frac{b}{a} \)
For our function transformation, finding \( A \) and \( \delta \) involves recognizing the coefficients of \( \sin \omega t \) and \( \cos \omega t \) as parts of a right triangle, where \( a \) and \( b \) are the opposite and adjacent sides. Solving this problem, we have \( a = 3 \), \( b = 4 \), yielding an amplitude \( A = 5 \) and phase \( \delta = \tan^{-1} \left(\frac{4}{3}\right) \).
This transformation of function forms demonstrates how trigonometric identities, complex exponentials, and properties like amplitude and phase seamlessly integrate to simplify and express complex functions.

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Most popular questions from this chapter

The definition of the dipole moment, \(\mathbf{D}=\sum q_{i} \mathbf{r}_{i}\), involves the vector \(\mathbf{r}_{i}\) stretching from the origin of our coordinate system out to the charge \(q_{i}\). There are clearly cases where this causes the dipole moment to be dependent on the choice of coordinate system. For instance, if there is only one charge, then we could make the dipole moment equal zero if we chose the origin to be right on top of the charge, or nonzero if we put the origin somewhere else. (a) Make up a numerical example with two charges of equal magnitude and opposite sign. Compute the dipole moment using two different coordinate systems that are oriented the same way, but differ in the choice of origin. Comment on the result. (b) Generalize the result of part a to any pair of charges with equal magnitude and opposite sign. This is supposed to be a proof for any arrangement of the two charges, so don't assume any numbers. (c) Generalize further, to \(n\) charges.

In a certain region of space, the electric field is constant (i.e., the vector always has the same magnitude and direction). For simplicity, assume that the field points in the positive \(x\) direction. (a) Use Gauss's law to prove that there is no charge in this region of space. This is most easily done by considering a Gaussian surface consisting of a rectangular box, whose edges are parallel to the \(x\), \(y\), and \(z\) axes. (b) If there are no charges in this region of space, what could be making this electric field?

A dipole has a midplane, i.e., the plane that cuts through the dipole's center, and is perpendicular to the dipole's axis. Consider a two-charge dipole made of point charges \(\pm q\) located at \(z=\pm \ell / 2\). Use approximations to find the field at a distant point in the midplane, and show that its magnitude comes out to be \(k D / R^{3}\) (half what it would be at a point on the axis lying an equal distance from the dipole).

A solid cylinder of radius \(b\) and length \(\ell\) is uniformly charged with a total charge \(Q .\) Find the electric field at a point at the center of one of the flat ends.

(a) At time \(t=0\), a positively charged particle is placed, at rest, in a vacuum, in which there is a uniform electric field of magnitude \(E\). Write an equation giving the particle's speed, \(v\), in terms of \(t, E\), and its mass and charge \(m\) and \(q\). (answer check available at lightandmatter.com) (b) If this is done with two different objects and they are observed to have the same motion, what can you conclude about their masses and charges? (For instance, when radioactivity was discovered, it was found that one form of it had the same motion as an electron in this type of experiment.)

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