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A transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s}\) . Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement, (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\)for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverse velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\) .

Short Answer

Expert verified
(a) Frequency: 20 Hz, Angular Frequency: 40π rad/s, Wave Number: 10π/9 rad/m. (b) Wave function: y(x, t) = 0.0025 sin((10π/9)x - 40πt). (c) y(t) = -0.0025 sin(40πt) at the origin. (d) y(t) = 0.0025 sin((4.5π/3) - 40πt) at x = 1.35 m. (e) Maximum transverse velocity: 0.1π m/s. (f) At x = 1.35 m, t = 0.0625 s: displacement = -0.00125 m, velocity ≈ -0.0545 m/s.

Step by step solution

01

Calculate the Frequency

The wave speed and wavelength are given, so we can use the formula: \[ f = \frac{v}{\lambda} \]where \( f \) is the frequency, \( v = 36.0 \, \mathrm{m/s} \) is the speed of the wave, and \( \lambda = 1.80 \, \mathrm{m} \) is the wavelength. Substituting the values, we get:\[ f = \frac{36.0 \, \mathrm{m/s}}{1.80 \, \mathrm{m}} = 20 \, \mathrm{Hz} \]
02

Calculate the Angular Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by the relation: \[ \omega = 2\pi f \] Substituting \( f = 20 \, \mathrm{Hz} \), we get: \[ \omega = 2\pi \times 20 = 40\pi \, \mathrm{rad/s} \]
03

Calculate the Wave Number

The wave number \( k \) can be calculated using the formula: \[ k = \frac{2\pi}{\lambda} \] where \( \lambda = 1.80 \, \mathrm{m} \). Thus:\[ k = \frac{2\pi}{1.80} = \frac{10\pi}{9} \, \mathrm{rad/m} \]
04

Write the Wave Function

The general form of the wave function for a sinusoidal wave traveling to the right is: \[ y(x, t) = A \sin(kx - \omega t) \] Where \( A = 2.50 \, \mathrm{mm} = 0.0025 \, \mathrm{m} \), \( k = \frac{10\pi}{9} \, \mathrm{rad/m} \), and \( \omega = 40\pi \, \mathrm{rad/s} \). Thus,\[ y(x, t) = 0.0025 \sin\left( \frac{10\pi}{9} x - 40\pi t \right) \]
05

Function at the Origin

For a particle at the origin \( x = 0 \), the wave function becomes: \[ y(0, t) = 0.0025 \sin(-40\pi t) \] As \( \sin(-\theta) = -\sin(\theta) \), we have:\[ y(0, t) = -0.0025 \sin(40\pi t) \]
06

Function at 1.35 m

For a particle at \( x = 1.35 \, \mathrm{m} \), the wave function is: \[ y(1.35, t) = 0.0025 \sin\left( \frac{10\pi}{9} \times 1.35 - 40\pi t \right) \]Calculate:\[ \frac{10\pi}{9} \times 1.35 = \frac{4.5\pi}{3} \]Thus,\[ y(1.35, t) = 0.0025 \sin\left( \frac{4.5\pi}{3} - 40\pi t \right) \]
07

Maximum Transverse Velocity

The maximum transverse velocity \( v_{\text{max}} \) is given by the product of the amplitude and the angular frequency:\[ v_{\text{max}} = A\omega = 0.0025 \times 40\pi = 0.1\pi \, \mathrm{m/s} \]
08

Displacement and Velocity at Specific Time

For the particle at \( x = 1.35 \, \mathrm{m} \) and \( t = 0.0625 \, \mathrm{s} \):1. Calculate displacement: \[ y(1.35, 0.0625) = 0.0025 \sin\left( \frac{4.5\pi}{3} - 40\pi \times 0.0625 \right) \] \[ y(1.35, 0.0625) = 0.0025 \sin\left( \frac{4.5\pi}{3} - 2.5\pi \right) \] \[ y(1.35, 0.0625) = 0.0025 \sin\left( -\frac{\pi}{6} \right) \] \[ y(1.35, 0.0625) = -0.0025 \times \frac{1}{2} = -0.00125 \, \mathrm{m} \]2. Calculate velocity: \[ v(1.35, 0.0625) = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t) \] \[ v(1.35, 0.0625) = -0.1\pi \cos\left( \frac{4.5\pi}{3} - 2.5\pi \right) \] \[ v(1.35, 0.0625) = -0.1\pi \cos\left( -\frac{\pi}{6} \right) \] \[ v(1.35, 0.0625) = -0.1\pi \times \frac{\sqrt{3}}{2} \approx -0.0545 \, \mathrm{m/s} \]

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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 concept in wave mechanics. They occur when the motion of the medium's particles is perpendicular to the direction of wave propagation. In our example, the wave is traveling along a stretched string. Imagine swings on a playground where the swings move up and down while the energy travels left to right.
Transverse waves are characterized by their amplitude, wavelength, frequency, and speed. The amplitude is the maximum distance a particle moves from its rest position as the wave travels. Here, it's 2.50 mm. The wavelength is the distance between two consecutive points in phase, such as two peaks. With a wavelength of 1.80 m, you can imagine this as the distance between crests in a wave on a string.
Transverse waves can be visualized as a sequence of peaks and troughs moving in a specific direction, transporting energy without the physical transport of matter across the medium.
Wave Function
The wave function is a crucial concept that describes the displacement of points on the medium as the wave passes through them. For sinusoidal waves on a string, the wave function is generally expressed in terms of sine or cosine functions.
The wave function for the given transverse wave is expressed as \( y(x, t) = A \sin(kx - \omega t) \). Here, \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency. This equation describes how the wave's displacement changes with position \( x \) and time \( t \). It's a snapshot of the wave at any given moment.
  • The term \( kx \) represents the spatial variation of the wave.
  • The term \( \omega t \) accounts for the temporal changes.
This equation allows you to predict the position of any point on the wave at any given time.
Angular Frequency
Angular frequency is a core concept when dealing with waves. It provides a measure of how quickly the wave oscillates in radians per second. For a sinusoidal wave, the angular frequency \( \omega \) is given by \( \omega = 2\pi f \), where \( f \) is the frequency of the wave.
In our exercise, the frequency was calculated using the wave speed and wavelength, resulting in \( f = 20\, \mathrm{Hz} \). Substituting this frequency into the formula gives \( \omega = 40\pi \mathrm{rad/s} \). Angular frequency helps to describe not just how many cycles occur per second but how much angle is swept during these cycles.
  • This parameter can translate into how quickly points on the wave move back and forth.
  • It essentially 'rotates' the unit circle as the wave advances, crucial for predicting wave behavior.
Understanding \( \omega \) allows you to grasp the rhythmic nature of wave motion on a deeper level.
Wave Number
Wave number \( k \) is another key parameter in wave mechanics, characterizing the spatial aspect of the wave. It refers to the number of wave cycles per unit distance and is closely linked with the wavelength \( \lambda \). It is expressed as \( k = \frac{2\pi}{\lambda} \).
For the given wave with a wavelength of 1.80 m, its wave number is calculated as \( k = \frac{10\pi}{9} \mathrm{rad/m} \). This number tells us how many radians the phase of the wave shifts per meter.
  • The larger the wave number, the more cycles fit into a unit length, indicating tighter wave crests.
  • This is crucial for understanding the spatial properties of waves, such as how packed they are or how quickly they oscillate over a distance.
Comprehending the wave number enables deeper insight into the spatial frequency of waves, complementing the understanding of their temporal frequency.

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Most popular questions from this chapter

Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 \(\mathrm{m} / \mathrm{s}\) and a frequency of 240 \(\mathrm{Hz}\) . The amplitude of the standing wave at an antinode is 0.400 \(\mathrm{cm}\) . (a) Calculate the amplitude at points on the string a distance of (i) \(40.0 \mathrm{cm} ;\) (ii) \(20.0 \mathrm{cm} ;\) and (iii) 10.0 \(\mathrm{cm}\) from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

A \(1.50-\mathrm{m}\) string of weight 1.25 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . When you pluck the string slightly, the waves traveling up the string obey the equation $$ y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-2730 \mathrm{s}^{-1} t\right) $$ (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\) ? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

The portion of the string of a certain musical instrument between the bridge and upper end of the finger board (that part of the string that is free to vibrate) is 60.0 \(\mathrm{cm}\) long. and this length of the string has mass 2.00 g. The string sounds an \(\mathrm{A}_{4}\) note \((440 \mathrm{Hz})\) when played. (a) Where must the player put a finger (what distance \(x\) from bridge) to play \(\mathbf{a}\) Ds note \((587 \quad \mathrm{Hz}) ? \quad\) (See Fig. 15.36 . For both the A_{4} and \(D_{5}\) notes, the string vibrates in its fundamental mode. (b) Without retuning, is it possible to play a G \(_{4}\) note \((392 \mathrm{Hz})\) on this string? Why or why not?

Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 \(\mathrm{Hz}\) to about 20.0 \(\mathrm{kHz}\) . (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart wonld the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meterstick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 \(\mathrm{m} / \mathrm{s}\) . How far apart would the dots be in each set? Could you readily measure their separation with a meterstick?

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