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Chapter 15: Problem 25

A sinusoidal wave traveling in the positive \(x\) -direction has a wavelength of \(12 \mathrm{~cm},\) a frequency of \(10.0 \mathrm{~Hz},\) and an amplitude of \(10.0 \mathrm{~cm}\). The part of the wave that is at the origin at \(t=0\) has a vertical displacement of \(5.00 \mathrm{~cm} .\) For this wave, determine the a) wave number, d) speed, b) period, e) phase angle, and c) angular frequency, f) equation of motion.

Short Answer

Expert verified
Based on the given information and step-by-step solutions, we found the following properties of the sinusoidal wave: 1. Wave number (k) = 52.36 m⁻¹ 2. Period (T) = 0.1 s 3. Angular frequency (ω) = 20π rad/s 4. Speed (v) = 1.2 m/s 5. Phase angle (φ) = 30° Finally, the equation of motion for the given wave is: y(x, t) = 10.0 cm sin(52.36 m⁻¹x - (20π rad/s)t + 30°)

Step by step solution

01

Wave number

The wave number (denoted by \(k\)) is the spatial frequency of the wave. To find the wave number, we can use the following equation: $$k = \frac{2\pi}{\lambda}$$ Here, \(\lambda\) is the wavelength which is given as \(12 cm\). So, let's calculate the wave number: $$k = \frac{2\pi}{12\ \text{cm}} = \frac{2\pi}{0.12\ \text{m}} \approx 52.36\ \text{m}^{-1}$$
02

Period

The period (denoted by \(T\)) is the time taken for one complete oscillation. It is related to the frequency (denoted by \(f\)) as follows: $$T = \frac{1}{f}$$ Here, the frequency is given as \(10.0 Hz\). So, let's calculate the period: $$T = \frac{1}{10.0\ \text{Hz}} = 0.1\ \text{s}$$
03

Angular frequency

The angular frequency (denoted by \(\omega\)) is related to the frequency by the following equation: $$\omega = 2\pi f$$ We have the frequency (\(10.0 Hz\)). Let's calculate the angular frequency: $$\omega = 2\pi (10.0\ \text{Hz}) = 20\pi\ \text{rad/s}$$
04

Speed

The wave speed (denoted by \(v\)) is the speed at which the wave travels in the medium. It can be found using the following equation: $$v = \lambda f$$ We have the wavelength (\(12 cm\)) and the frequency (\(10.0 Hz\)). Let's calculate the wave speed: $$v = (12\ \text{cm})(10.0\ \text{Hz}) = 1.2\ \text{m/s}$$
05

Phase angle

To find the phase angle (denoted by \(\phi\)), we need to consider the vertical displacement at the origin when \(t=0\). Let's denote the vertical displacement as \(y\). The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ At \(t=0\) and \(x=0\), \(y(0,0) = 5.00\ \text{cm}\). Let's substitute the values and solve for \(\phi\): $$5.00\ \text{cm} = (10.0\ \text{cm})\sin(\phi)$$ $$\phi = \arcsin\left(\frac{1}{2}\right)$$ $$\phi = 30^{\circ}$$
06

Equation of motion

Now we have all the necessary parameters to write the equation of motion for the given wave. The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ Substitute the values we found for \(A\), \(k\), \(\omega\), and \(\phi\): $$y(x, t) = 10.0\ \text{cm}\sin\left(52.36\ \text{m}^{-1}x - (20\pi\ \text{rad/s})t + 30^{\circ}\right)$$

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