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The equation describing a transverse wave on a string is $$y(x, t)=(1.50 \mathrm{~mm}) \sin \left[\left(157 \mathrm{~s}^{-1}\right) t-\left(41.9 \mathrm{~m}^{-1}\right) x\right]$$ Find (a) the wavelength, frequency, and amplitude of this wave, (b) the speed and direction of motion of the wave, and (c) the transverse displacement of a point on the string when \(t=0.100 \mathrm{~s}\) and at a position \(x=0.135 \mathrm{~m}\)

Short Answer

Expert verified
(a) \( \lambda = 0.150 \mathrm{~m} \), \( f = 25.0 \mathrm{~Hz} \), \( A = 1.50 \mathrm{~mm} \); (b) Speed = 3.75 m/s, Direction: positive x; (c) Displacement = -1.47 mm at \( t = 0.100 \mathrm{~s} \), \( x = 0.135 \mathrm{~m} \).

Step by step solution

01

Identify the Given Parameters

The wave equation is given as \( y(x, t)=(1.50 \mathrm{~mm}) \sin \left[\left(157 \mathrm{~s}^{-1}\right) t-\left(41.9 \mathrm{~m}^{-1}\right) x\right] \). Here, the amplitude \( A = 1.50 \mathrm{~mm} \), the angular frequency \( \omega = 157 \mathrm{~s}^{-1} \), and the wave number \( k = 41.9 \mathrm{~m}^{-1} \).
02

Find the Wavelength

The wavelength \( \lambda \) is related to the wave number \( k \) by the equation \( k = \frac{2\pi}{\lambda} \). Thus, \( \lambda = \frac{2\pi}{41.9} \approx 0.150 \mathrm{~m} \).
03

Find the Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by \( \omega = 2\pi f \). Thus, \( f = \frac{157}{2\pi} \approx 25.0 \mathrm{~Hz} \).
04

Confirm the Amplitude

The amplitude is directly given in the equation as \( A = 1.50 \mathrm{~mm} \). Hence, the amplitude of the wave is \( 1.50 \mathrm{~mm} \).
05

Calculate the Wave Speed

The wave speed \( v \) can be found using the relationship \( v = f \cdot \lambda \). Using the found values, \( v = 25.0 \mathrm{~Hz} \times 0.150 \mathrm{~m} = 3.75 \mathrm{~m/s} \).
06

Determine the Direction

The direction of the wave is determined by the sign in the argument of the sine function. The negative sign \(-\left(41.9 \mathrm{~m}^{-1}\right) x\) indicates the wave is traveling in the positive x-direction.
07

Calculate Transverse Displacement at Given Time and Position

Substitute \( x = 0.135 \mathrm{~m}\) and \( t = 0.100 \mathrm{~s} \) into the wave equation: \[y(0.135, 0.100) = 1.50 \sin\left(157 \times 0.100 - 41.9 \times 0.135\right) \approx 1.50 \sin(15.7 - 5.6565) \approx 1.50 \sin(10.0435) \approx -1.47 \mathrm{~mm}.\]

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Key Concepts

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

Transverse Waves
Transverse waves are a fundamental type of wave where the particles of the medium move perpendicular to the direction of the wave's travel. Imagine a rope being flicked up and down; as a result, the wave moves horizontally along the rope, while the particles of the rope move vertically. This distinct perpendicular motion characterizes transverse waves and differentiates them from other wave types, such as longitudinal waves where particle movement is parallel to wave direction.

In our problem, we consider a transverse wave traveling along a string. The variables in our wave equation, like amplitude and frequency, describe the wave’s characteristics, and understanding these helps us predict the wave’s motion. For instance, the amplitude (1.50 mm) suggests how far particles in the medium move from their rest position, showing the wave's energy level.
Wave Equation
The wave equation describes the displacement of a wave as it moves through a medium. For a transverse wave, the equation often takes the form: \[ y(x, t) = A \sin(\omega t - kx) \] where
  • \( y(x, t) \) is the wave's displacement at position \( x \) and time \( t \).
  • \( A \) is the amplitude, representing the maximum displacement.
  • \( \omega \) is the angular frequency, determining how many oscillations occur per second.
  • \( k \) is the wave number, related to the wavelength and direction of the wave.
The terms in the sine function determine the wave’s progression over time and space. The negative sign in \(\omega t - kx\) indicates the wave travels in the positive x-direction. This clear framework allows for predicting how the wave evolves, revealing information like speed and direction by substituting known values.
Wavelength
The wavelength \( \lambda \) of a wave is the distance over which the wave's shape repeats. It holds the key to understanding how many wave crests (or troughs) fit within a unit of distance. In our scenario, the relationship between the wave number \( k \) and the wavelength is given by: \[ k = \frac{2\pi}{\lambda} \] From this equation, we can rearrange to find the wavelength: \[ \lambda = \frac{2\pi}{k} \] Plugging the values, we determine that \( \lambda \approx 0.150\, \text{m} \).

A smaller wavelength indicates a denser packing of wave peaks and troughs, which often means higher frequency, and vice versa. The wavelength is crucial in applications like communication, where signals are transmitted in specific wave bands.
Angular Frequency
Angular frequency \( \omega \) is a measure of how quickly a wave oscillates and is central to understanding wave behavior. It is expressed in radians per second. The relationship between angular frequency and regular frequency \( f \) is straightforward: \[ \omega = 2\pi f \] Using this relationship, we calculate \( f \) from a given \( \omega \). In our case, \( \omega = 157\, \text{s}^{-1} \), which translates to a frequency \( f \approx 25.0 \) Hz, meaning 25 oscillations per second occur.

This concept is essential in fields such as acoustics and optics, where understanding how fast wave crests pass a certain point informs us about the energy and nature of the wave. Angular frequency ensures a concise description of this oscillatory motion, compactly expressing its periodic nature.

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Most popular questions from this chapter

Transverse waves are traveling on a long string that is under a tension of \(4.00 \mathrm{~N}\). The equation describing these waves is $$y(x, t)=(1.25 \mathrm{~cm}) \sin \left[\left(415 \mathrm{~s}^{-1}\right) t-\left(44.9 \mathrm{~m}^{-1}\right) x\right]$$ (a) Find the speed of the wave using the equation. (b) Find the lincar mass density of this string.

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