Chapter 7: Problem 6

Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance $s$ from the origin is the given function. $$s=3 \sin (2 t+\pi / 8)+3 \sin (2 t-\pi / 8)$$

Short Answer

Expert verified

Amplitude: 3√2, Period: π, Frequency: 1/π, Velocity Amplitude: 6√2.

Step by step solution

- Simplify the given function

First, simplify the given function using trigonometric identities. The given function is: \[ s = 3 \sin(2t + \frac{\pi}{8}) + 3 \sin(2t - \frac{\pi}{8}) \]Using the sum-to-product identities, \[ \sin A + \sin B = 2 \sin\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right) \]Set $ A = 2t + \frac{\pi}{8} $ and $ B = 2t - \frac{\pi}{8} $:\[ A+B = 4t \text{ and } A-B = \frac{2\pi}{8} = \frac{\pi}{4} \]So the function becomes:\[ s = 2 \cdot 3 \sin(2t) \cos(\frac{\pi}{4}) \]Since $ \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} $:\[ s = 6 \sin(2t) \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \sin(2t) \]

- Identify amplitude

The simplified function is \[ s = 3\sqrt{2} \sin(2t) \]The amplitude is the coefficient of the sine function, which is \[ 3\sqrt{2} \]

- Determine the period

The period of a sine function of the form $ \sin(bt) $ is given by $ \frac{2\pi}{b} $. Here, $ b = 2 $, so the period is:\[ T = \frac{2\pi}{2} = \pi \]

- Calculate frequency

The frequency is the reciprocal of the period. Since $ T = \pi $:\[ f = \frac{1}{T} = \frac{1}{\pi} \]

- Find the velocity amplitude

The velocity amplitude is the product of the amplitude and the angular frequency. The angular frequency $ \omega $ is given by $ b $ in the sine function $ \sin(bt) $.Here, $ b = 2 $. Hence,\[ A_v = Amplitude \times \omega = 3\sqrt{2} \times 2 = 6\sqrt{2} \]

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Key Concepts

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

Amplitude

In the context of harmonic motion, the amplitude refers to the maximum distance a particle moves from its equilibrium (rest) position. Amplitude gives us a measure of the strength or intensity of the motion. In the given function, we simplified it to $ s = 3\sqrt{2} \sin(2t) $. The amplitude is the coefficient of the sine function, which here is $ 3\sqrt{2} $. This value indicates the peak magnitude of the oscillation of the particle. So, when you see something like $ 3\sqrt{2} $, think of it as telling you how far the particle will move at its most extreme point from the center.

Period

To understand the period, imagine how long it takes for the particle to complete one full cycle of motion. The period is the duration it takes for the system to return to its initial position and velocity. For a sine function like $ \sin(bt) $, the period $ T $ is calculated using the formula: $ T = \frac{2\pi}{b} $. In our equation, $ b = 2 $, giving us a period $ T = \frac{2\pi}{2} = \pi $. This means that every $ \pi $ units of time, the particle completes a full cycle of its motion. It’s a neat way to determine the time frame of repetitive cycles in the motion.

Frequency

Frequency tells us how many cycles of the motion occur in one unit of time. It's the flip side of the period. If we know the period, we can calculate the frequency as the reciprocal of the period: $ f = \frac{1}{T} $. From our earlier calculation, that would be: $ f = \frac{1}{\pi} $. Frequency is essential for understanding how often the motion repeats within a certain time span. So, with a frequency of $ \frac{1}{\pi} $, it describes how many times per unit of time our particle goes through its full range of motion.

Velocity Amplitude

Velocity amplitude might sound complex, but it's just how fast the particle is moving at its maximum speed. To find it, we need to know the amplitude and the angular frequency ($ \omega $). The angular frequency here is denoted by $ b $ from our function $ \sin(bt) $. In this example, $ b = 2 $, so $ \omega = 2 $. The velocity amplitude is then the product of the amplitude and $ \omega $. Here, we get: $ A_v = Amplitude \times \omega = 3\sqrt{2} \times 2 = 6\sqrt{2} $. This tells us that the highest rate of change of the particle's position (how fast it can be) is $ 6\sqrt{2} $. This value is crucial for understanding the dynamic nature of our particle's motion.

Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance \(s\) from the origin is the given function. $$s=3 \sin (2 t+\pi / 8)+3 \sin (2 t-\pi / 8)$$