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Transverse waves on a string have wave speed 8.00 \(\mathrm{m} / \mathrm{s}\) , amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m} .\) The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a par-ticle at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) . (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{m}\) next has maximum upward displacement?

Short Answer

Expert verified
(a) \(f = 25.0 \, \text{Hz}\), \(T = 0.040 \, \text{s}\), \(k = 19.63 \, \text{m}^{-1}\). (b) \(y = 0.0700 \sin(19.63x + 157.08t + \frac{\pi}{2})\). (c) \(y = 0.048 \, \text{m}\). (d) \(\Delta t \approx 0.010 \, \text{s}\).

Step by step solution

01

Identify Given Data

We are given the wave speed \(v = 8.00 \, \mathrm{m/s}\), amplitude \(A = 0.0700 \, \text{m}\), and wavelength \(\lambda = 0.320 \, \text{m}\). The wave travels in the \(-x\) direction and at time \(t = 0\), the end of the string at \(x = 0\) has its maximum displacement.
02

Calculate Frequency

The frequency \( f \) of a wave is given by the formula \( f = \frac{v}{\lambda} \). Substituting the values, we have \( f = \frac{8.00}{0.320} = 25.0 \, \text{Hz} \).
03

Calculate Period

The period \( T \) is the reciprocal of the frequency: \( T = \frac{1}{f} = \frac{1}{25.0} = 0.040 \, \text{s} \).
04

Calculate Wave Number

The wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \). Substituting for \(\lambda\), \( k = \frac{2\pi}{0.320} \approx 19.63 \, \text{m}^{-1} \).
05

Write the Wave Function

The general form of a wave traveling in the \(-x\)-direction is \( y(x,t) = A \sin(kx + \omega t + \phi) \). Here, \( \omega = 2\pi f \approx 157.08 \, \text{rad/s} \). Since the wave is at a maximum positive displacement at \(x=0\) and \(t=0\), we have \(\phi = \frac{\pi}{2}\). The wave function becomes \( y(x,t) = 0.0700 \sin(19.63x + 157.08t + \frac{\pi}{2}) \).
06

Calculate Transverse Displacement at x=0.360 m, t=0.150 s

Substitute \(x = 0.360 \, \text{m}\) and \(t = 0.150 \, \text{s}\) into the wave function: \( y(0.360, 0.150) = 0.0700 \sin(19.63 \times 0.360 + 157.08 \times 0.150 + \frac{\pi}{2}) \). Calculating, \( y \approx 0.048 \, \text{m} \).
07

Determine Time for Next Maximum Upward Displacement

A point has maximum upward displacement when the sine function is \(1\). Starting from \(\theta = 19.63 \times 0.360 + 157.08 \times 0.150 + \frac{\pi}{2} = \theta_0\), find the next \(\theta\) where \(\sin(\theta) = 1\), which happens every period: \(\theta = \theta_0 + 2\pi\). Solve \(19.63 \times 0.360 + 157.08(t+0.150) + \frac{\pi}{2} = \theta_0 + 2\pi\). Solving for \(t\), \(t \approx 0.010 \, \text{s} \).

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Key Concepts

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

Wave Speed
Wave speed is a fundamental property of waves, describing how quickly they travel through a medium. It's represented by the symbol \( v \) and is expressed in meters per second (m/s). In our exercise, the transverse waves on the string have a given speed of 8.00 m/s.

Wave speed is related to two primary characteristics of a wave:
  • Wavelength (\( \lambda \)): The distance over which the wave's shape repeats. For our wave, \( \lambda = 0.320 \, \text{m} \).
  • Frequency (\( f \)): The number of cycles per second, measured in Hertz (Hz).
These properties are linked through the formula for wave speed:\[v = f \times \lambda\\]Using this relation, you can calculate any of the properties if the other two are known. Understanding wave speed allows us to predict how a wave will propagate through a medium over time.
Frequency Calculation
To calculate the frequency of a wave, you can use the formula:\[f = \frac{v}{\lambda}\\]This equation demonstrates that frequency is the wave speed divided by the wavelength. In our specific exercise, substituting the known values gives:\[f = \frac{8.00}{0.320} = 25.0 \, \text{Hz}\\]Frequency tells us how many times a wave oscillates each second. It is a key variable that not only defines a wave's behavior, but also influences other characteristics like the wave's period.
  • Period (\( T \)): The time it takes for one full cycle of the wave, calculated as the reciprocal of the frequency: \( T = \frac{1}{f} = 0.040 \, \text{s} \). The period is inversely proportional to frequency; hence, a high frequency means a short period and vice versa.
Wave Function
The wave function describes the shape and behavior of a wave at any given point in space and time. It provides a complete mathematical model for what the wave does. For a wave traveling in the -x direction, the general form is:\[y(x,t) = A \sin(kx + \omega t + \phi)\\]Here's what each term signifies:
  • \( A \): Amplitude, the maximum displacement from the equilibrium position. Here \( A = 0.0700 \, \text{m} \).
  • \( k \): Wave number, calculated by \( k = \frac{2\pi}{\lambda} \approx 19.63 \, \text{m}^{-1} \). It measures how many wave cycles occur in a unit distance.
  • \( \omega \): Angular frequency, given by \( \omega = 2\pi f \approx 157.08 \, \text{rad/s} \).
  • \( \phi \): Phase constant, which determines the wave's initial phase. In this exercise, it's set to \( \phi = \frac{\pi}{2} \) since the particle is at its highest point at \( t=0 \).
The specific wave function for our exercise becomes:\[y(x,t) = 0.0700 \sin(19.63x + 157.08t + \frac{\pi}{2})\\]This function can be used to compute the wave's position at any location and time, giving deep insight into its dynamic behavior.
Transverse Displacement
Transverse displacement refers to the vertical shift of a point on the string from its equilibrium position as the wave passes. Calculating this involves substituting the desired values for position \( x \) and time \( t \) into the wave function derived earlier.

For instance, in this specific problem, you need to find the displacement at \( x = 0.360 \, \text{m} \) and \( t = 0.150 \, \text{s} \):\[y(0.360, 0.150) = 0.0700 \sin(19.63 \times 0.360 + 157.08 \times 0.150 + \frac{\pi}{2})\\]By performing the calculations, we find that the displacement at these coordinates is approximately 0.048 m.

This tells us how much a point on the string has moved from its resting position at that specific place and time. Understanding transverse displacement helps in visualizing wave motion and examining how different points on a medium move in response to a wave's energy. This is crucial in fields like engineering, meteorology, and even music, where wave behavior is fundamental.

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Most popular questions from this chapter

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 dis- tance 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).

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