Problem 10 You are given a complex function... [FREE SOLUTION]

Chapter 7: Problem 10

You are given a complex function $z=f(t) .$ In each case, show that a particle wose 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. $$ z=-4 e^{i(2 t+3 k)} $$

Short Answer

Expert verified

Both x and y undergo SHM with amplitude 4, period $ \pi $, frequency $ \frac{1}{\pi} $, and velocity amplitude 8 units/s.

Step by step solution

Express the complex function

Given the complex function $z = -4 e^{i(2t + 3k)}$, we use Euler's formula $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ to rewrite it in trigonometric form: \[-4 e^{i(2t + 3k)} = -4( \cos(2t + 3k) + i \sin(2t + 3k)).\]

Simplify the expression

Simplify the expression by distributing the -4: \[ z = -4 \cos(2t + 3k) - 4i \sin(2t + 3k). \]

Identify the real and imaginary parts

Identify the real and imaginary components: $ x = \operatorname{Re}(z) = -4 \cos(2t + 3k) $ and $ y = \operatorname{Im}(z) = -4 \sin(2t + 3k) $.

Show simple harmonic motion for x

The coordinate $ x = -4 \cos(2t + 3k) $ represents simple harmonic motion (SHM) because it is in the form $ A \cos(Bt + \phi) $. Here, the amplitude $ A = 4 $, the angular frequency $ B = 2 $, and phase shift $ \phi = 3k $.

Show simple harmonic motion for y

Similarly, the coordinate $ y = -4 \sin(2t + 3k) $ is also in the form of SHM $ A \sin(Bt + \phi) $. Here, the amplitude is $ A = 4 $, the angular frequency is $ B = 2 $, and the phase shift is $ \phi = 3k $.

Determine amplitude

For both parts (a) and (b), the amplitude of the motion is the constant factor before the trigonometric function, which is $ A = 4 $.

Calculate the period

The period $ T $ is calculated by the formula $ T = \frac{2\pi}{B} $. Given that $ B = 2 $, \[ T = \frac{2\pi}{2} = \pi. \]

Calculate the frequency

The frequency $ f $ is the reciprocal of the period $ T $, so $ f = \frac{1}{T} = \frac{1}{\pi} $.

Determine velocity amplitude

The amplitude of the velocity can be found by differentiating the position function. For $ x $, $ v_x = \frac{dx}{dt} = 4 \cdot 2 \sin(2t + 3k) = 8 \sin(2t + 3k) $, so the velocity amplitude is $ 8 \text{units/s} $. Similarly for $ y $, $ v_y = \frac{dy}{dt} = -4 \cdot 2 \cos(2t + 3k) = -8 \cos(2t + 3k) $, so the velocity amplitude is also $ 8 \text{units/s} $.

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Key Concepts

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

complex functions

Complex functions are mathematical expressions involving complex numbers. Complex numbers have a real component and an imaginary component. We represent them as z = x + yi, where x is the real part and yi is the imaginary part. Working with complex functions often involves understanding these two components separately.
In this exercise, the given complex function is z = -4 e^{i(2t + 3k)}. This notation indicates the function depends on the variable t and constants, and has both magnitude and direction.

Euler's formula

Euler's formula is a key tool when dealing with complex functions. It states that for any real number 胃, e^{i胃} = cos(胃) + i sin(胃). This formula helps convert complex exponential functions into trigonometric functions, which are often easier to manipulate.
For our problem, Euler's formula allows us to rewrite -4 e^{i(2t + 3k)} as -4 (cos(2t + 3k) + i sin(2t + 3k)). This separation of the function into real and imaginary parts makes further calculations more straightforward.

simple harmonic motion

Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. In this context, SHM appears as sine or cosine functions.
For the coordinates in our exercise: x = -4 cos(2t + 3k) and y = -4 sin(2t + 3k), both exhibit SHM because they are in the forms A cos(Bt + 蠁) and A sin(Bt + 蠁) respectively. This confirms that the motions of x and y are both simple harmonic.

amplitude

Amplitude in the context of SHM refers to the maximum extent of the oscillation from its mean position. It is the constant factor before the trigonometric function.
In our problem, for both x and y, the amplitude is the absolute value of -4, which is 4. This tells us that the particle oscillates 4 units away from its mean position in both coordinates.

angular frequency

Angular frequency, often denoted as 蠅, describes how quickly the oscillations of SHM occur. It is related to the period and frequency of the motion and appears in the arguments of the sine and cosine functions.
From the given function, the argument 2t + 3k shows that the angular frequency (B) is 2. This value helps determine the overall period and frequency of the motion.

velocity amplitude

Velocity amplitude is the peak velocity that a particle reaches during SHM. It is determined by differentiating the position function concerning time.
For x: differentiate -4 cos(2t + 3k) to get 8 sin(2t + 3k), thus the velocity amplitude is 8 units/s. Similarly for y: differentiate -4 sin(2t + 3k) to get -8 cos(2t + 3k), leading to the same amplitude of 8 units/s. This shows that the maximum speed of the particle is consistent across both coordinates.

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