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A continuous sinusoidal longitudinal wave is sent along a very long coiled spring from an attached oscillating source. The wave travels in the negative direction of an \(x\) axis; the source frequency is \(25 \mathrm{~Hz}\); at any instant the distance between successive points of maximum expansion in the spring is \(24 \mathrm{~cm}\); the maximum longitudinal displacement of a spring particle is \(0.30 \mathrm{~cm} ;\) and the particle at \(x=0\) has zero displacement at time \(t=0\). If the wave is written in the form \(s(x, t)=s_{m} \cos (k x \pm \omega t)\), what are (a) \(s_{m}\), (b) \(k\), (c) \(\omega\), (d) the wave speed, and (e) the correct choice of sign in front of \(\omega ?\) Incident

Step by step solution

01

Determine the Amplitude ( $s_{m}$)

The maximum longitudinal displacement of a spring particle is given as 0.30 cm. Thus, the amplitude of the wave is directly the maximum displacement.So, \(s_{m} = 0.30 \text{ cm} = 0.0030 \text{ m}\)

02

Calculate the Wave Number ($k$)

The distance between successive points of maximum expansion in the spring is the wavelength \(\lambda\). This is given as 24 cm. First, convert this into meters:\[\lambda = 24 \text{ cm} = 0.24 \text{ m}\]The wave number \(k\) is calculated using the formula:\[k = \frac{2\pi}{\lambda}\]Substitute the wavelength:\[k = \frac{2\pi}{0.24} \approx 26.18 \text{ m}^{-1}\]

03

Calculate the Angular Frequency ($ $\omega$)

The source frequency \(f\) is given as 25 Hz. The angular frequency \(\omega\) is related to the frequency by the formula:\[\omega = 2\pi f\]Substitute the known frequency:\[\omega = 2\pi \times 25 \approx 157.08 \text{ rad/s}\]

04

Determine the Wave Speed (v)

The wave speed can be determined using the relationship between speed, frequency, and wavelength:\[v = f \times \lambda\]Substituting the known values:\[v = 25 \times 0.24 = 6.0 \text{ m/s}\]

05

Determine the Correct Sign in the Equation

The wave is traveling in the negative direction of the x-axis, which means we should choose the positive sign for the angular frequency term in the wave equation. Thus, the wave function has the form:\[s(x, t) = s_{m} \cos(kx + \omega t)\]

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Key Concepts

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

Amplitude

In wave properties, amplitude is a measure of the wave's maximum displacement from its rest position. This can be visualized as the height of the wave crest or the depth of the wave trough. It essentially represents how strong or intense a wave is. The amplitude is important because it directly affects the energy carried by a wave. For the given problem, the amplitude, denoted as \( s_m \), is the maximum longitudinal displacement recorded, which is \( 0.30 \) cm. It's crucial to convert units appropriately, and thus it becomes \( 0.0030 \) m when converted to meters. Proper unit conversion ensures accuracy in wave calculations.

Wave Number

Wave number is a fundamental characteristic of a wave that relates to its wavelength. It provides a measure of the number of wave cycles in a specific unit length. The wave number is symbolized as \( k \), and it's defined mathematically as \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. The wavelength in this scenario is \( 24 \) cm, equivalent to \( 0.24 \) m when converted to meters. Substituting this value into the formula gives a wave number of approximately \( 26.18 \text{ m}^{-1} \). The unit \( \text{m}^{-1} \) (inverse meters) highlights how many wave cycles fit in one meter.

Angular Frequency

Angular frequency is closely related to a wave's temporal rate of oscillation and denotes how many radians a wave covers per second. It is represented by the symbol \( \omega \) and is calculated using the formula \( \omega = 2\pi f \), where \( f \) is the frequency of the wave. For the problem at hand, the source frequency is \( 25 \) Hz, so substituting into the formula gives \( \omega \approx 157.08 \text{ rad/s} \). This measure essentially reveals how quickly the wave oscillates in time, with "rad/s" indicating radians per second.

Wave Speed

Wave speed is an essential property that informs us how fast the wave travels through a medium. It's the speed at which the wave's energy is transmitted. The relationship connecting wave speed \( v \), frequency \( f \), and wavelength \( \lambda \) is given by \( v = f \times \lambda \). With the provided frequency of \( 25 \) Hz and wavelength \( 0.24 \) m, the wave speed can be calculated as \( v = 25 \times 0.24 = 6.0 \text{ m/s} \). This result shows that each wave crest moves along the spring at \( 6.0 \text{ m/s} \). Understanding wave speed is important in many practical applications, such as sound waves traveling through the air or light waves in optical fibers.