Problem 8 You are given a complex function... [FREE SOLUTION]

Chapter 7: Problem 8

You are given a complex function $z=f(t) .$ In each case, show that a particle whose coordinate is (a) $x=\operatorname{Re} z,$ (b) $y=\operatorname{Im} z$ is undergoing simple harmonic motion, and find the amplitude, period, frequency, and velocity amplitude of the motion. $$\quad z=2 e^{-i t / 2}$$

Short Answer

Expert verified

Amplitude = 2, Period = 4蟺, Frequency = 1/(4蟺), Velocity Amplitude = 1.

Step by step solution

- Analyze the given complex function

The given function is $ z = 2 e^{-i t / 2} $. This can be rewritten using Euler's formula $ e^{i\theta} = \cos \theta + i \sin \theta $. Therefore, $ z = 2 ( \cos(-t/2) + i \sin(-t/2) ) $. Since cosine is an even function $ \cos(-\theta) = \cos(\theta) $ and sine is an odd function $ \sin(-\theta) = -\sin(\theta) $, the function simplifies to $ z = 2 \cos(t/2) - 2i \sin(t/2) $.

- Separate the real and imaginary parts

Identify the real and imaginary parts of the function: $ x = \operatorname{Re}(z) = 2 \cos(t/2) $ and $ y = \operatorname{Im}(z) = -2 \sin(t/2) $.

- Show that x is undergoing simple harmonic motion

Simple harmonic motion (SHM) can be characterized by a function of the form $ x(t) = A \cos( \omega t + \phi ) $. Here, $ x = 2 \cos(t/2) $, which matches the SHM form with amplitude $ A = 2 $ and angular frequency $ \omega = 1/2 $.

- Show that y is undergoing simple harmonic motion

Similarly, consider $ y(t) = -2 \sin(t/2) $. This can also represent SHM because $ -\sin(\theta) $ is simply a phase-shifted cosine function. Therefore, the amplitude $ A = 2 $ and angular frequency $ \omega = 1/2 $ for $ y(t) $ as well.

- Determine the amplitude, period, frequency, and velocity amplitude

For both $ x(t) $ and $ y(t) $: - **Amplitude (A)**: 2 - **Angular Frequency ($ \omega $)**: 1/2 - **Period (T)**: $ T = \frac{2\pi}{\omega} = 4\pi $ - **Frequency (f)**: $ f = \frac{1}{T} = \frac{1}{4\pi} $ - **Velocity amplitude (v_{max})**: Given by $ A \omega $, which is $ 2 \cdot \frac{1}{2} = 1 $.

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Key Concepts

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

Euler's Formula

Euler's formula, given by $ e^{i\theta} = \cos \theta + i \sin \theta $, serves as a crucial tool in understanding complex harmonic motion. This formula allows the transformation of exponential functions involving imaginary numbers into trigonometric functions. By applying Euler's formula to the given complex function $ z = 2 e^{-i t / 2} $, it can be rewritten as $ z = 2 ( \cos(-t/2) + i \sin(-t/2) ) $. This simplification helps us separate the real and imaginary parts, which are important for analyzing motion in terms of simple harmonic motion.

Simple Harmonic Motion

Simple harmonic motion (SHM) describes a type of periodic oscillatory motion where the restoring force is directly proportional to the displacement. The general form of SHM can be expressed as $ x(t) = A \cos( \omega t + \phi ) $. In this exercise, extracting the real part $ x = 2 \cos(t/2) $ and imaginary part $ y = -2 \sin(t/2) $ of the function $ z = 2 e^{-i t / 2} $ reveals that both represent SHM. The amplitude $ A $, angular frequency $ \omega $, and other characteristics confirm that the particle demonstrates SHM along both axes.

Angular Frequency

Angular frequency $ \omega $ is a measure of how quickly an object moves through its cycle of motion. It is directly related to the period and frequency of SHM. For the function $ x = 2 \cos(t/2) $, the angular frequency $ \omega $ is $ 1/2 $. Angular frequency is measured in radians per second and is crucial for calculating the period and frequency of the motion.

Amplitude

Amplitude $ A $ represents the maximum displacement from the equilibrium position in either direction during SHM. It is a measure of the motion's 'width.' For both $ x(t) = 2 \cos(t/2) $ and $ y(t) = -2 \sin(t/2) $, the amplitude is 2. This value signifies the maximum extent of the particle's deviation from the mean position.

Period

The period $ T $ of a simple harmonic oscillator is the time it takes to complete one full cycle of motion. It is inversely related to the angular frequency $ \omega $ by the formula $ T = \frac{2\pi}{\omega} $. Given the angular frequency $ \omega = 1/2 $, the period for our motion is $ T = 4\pi $ seconds. This means the particle returns to its initial position after every $ 4\pi $ seconds.

Frequency

Frequency $ f $ represents the number of cycles completed per second in SHM. It is the reciprocal of the period $ T $, expressed as $ f = \frac{1}{T} $. For our case, with $ T = 4\pi $, the frequency is $ f = \frac{1}{4\pi} $. Hence, the particle completes approximately $ 0.0796 $ cycles per second.

Velocity Amplitude

Velocity amplitude represents the maximum speed achieved during SHM and is calculated by the product of the amplitude $ A $ and the angular frequency $ \omega $. For both $ x(t) = 2 \cos(t/2) $ and $ y(t) = -2 \sin(t/2) $, we have $ A = 2 $ and $ \omega = 1/2 $. Therefore, the velocity amplitude is $ v_{max} = A \omega = 2\cdot (1/2) = 1 $. This indicates that the particle reaches its maximum speed of 1 unit per second during its cycle.

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