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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).

Short Answer

Expert verified
Amplitude is 0.4 cm at 20 cm, and 0.283 cm at 10 cm. Time for largest displacement change is 0.00208 s at all points. Maximum velocity and acceleration are 6.03 m/s and 9075 m/s², respectively, at antinodes.

Step by step solution

01

Understand Harmonic Motion

First, understand that the string is vibrating in its third harmonic, which means there are three antinodes and three nodes on the string. This is important to determine where along the string maximum displacement occurs.
02

Equation of Motion for Standing Waves

The amplitude at any point on a standing wave can be determined using the equation: \[ A(x) = A_0 \sin(kx) \]where \( A_0 = 0.004 \text{ m} \) is the maximum amplitude, \( k \) is the wave number, and \( x \) is the position along the string.
03

Calculate Wave Number

Calculate the wave number \( k \) using the formula: \[ k = \frac{2\pi}{\lambda} \] First, find the wavelength, \( \lambda \), using the relationship between speed, frequency, and wavelength: \[ \lambda = \frac{v}{f} = \frac{192 \ \mathrm{m/s}}{240 \ \mathrm{Hz}} = 0.8 \ \mathrm{m} \]Then, \[ k = \frac{2\pi}{0.8} = 7.85 \ \mathrm{rad/m} \]
04

Calculate Amplitude at 40 cm

Substitute \( x = 0.4 \ \mathrm{m} \) along with previously calculated values into the equation: \[ A(0.4) = 0.004 \sin(7.85 \times 0.4) = 0.004 \sin(3.14) = 0 \ m \]Thus, the amplitude at \( \text{40 cm} \) is zero, which corresponds to a node.
05

Calculate Amplitude at 20 cm

For \( x = 0.2 \ \mathrm{m} \):\[ A(0.2) = 0.004 \sin(7.85 \times 0.2) = 0.004 \sin(1.57) = 0.004 \ m \]Thus, the amplitude at \( \text{20 cm} \) is \( \text{0.004 m or 0.4 cm}\).
06

Calculate Amplitude at 10 cm

For \( x = 0.1 \ \mathrm{m} \):\[ A(0.1) = 0.004 \sin(7.85 \times 0.1) = 0.004 \sin(0.785) = 0.00283 \ m \]Thus, the amplitude at \( \text{10 cm} \) is approximately \( \text{0.283 cm} \).
07

Calculate Time for Maximum Displacement Change

The time taken from largest upward to largest downward displacement is half the period of oscillation, calculated as: \[ T = \frac{1}{f} = \frac{1}{240} = 0.004167 \, \mathrm{s} \]Thus, time taken: \( \frac{T}{2} = 0.00208 \ \mathrm{s} \). This is the same at all points.
08

Maximum Transverse Velocity

The maximum transverse velocity \( v_{max} \) is given by: \[ v_{max} = A_0 \cdot \omega \] Where \( \omega = 2\pi f \), calculate: \[ \omega = 2\pi \times 240 = 1507 \ \mathrm{rad/s} \]Therefore, \[ v_{max} = 0.004 \times 1507 = 6.03 \ \mathrm{m/s} \]
09

Maximum Transverse Acceleration

The maximum transverse acceleration \( a_{max} \) is: \[ a_{max} = A_0 \cdot \omega^2 \] Using the calculated \( \omega \): \[ a_{max} = 0.004 \times (1507)^2 = 9075 \ \mathrm{m/s^2} \] This is the maximum at the antinode. Use amplitude found at points for specific calculations.

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Key Concepts

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

Harmonic Motion
In the context of vibrating strings, harmonic motion refers to the repetitive oscillation occurring when the string vibrates. For a string fixed at both ends, like our example, the harmonics correspond to the specific modes of vibration. Each mode has a specific number of nodes and antinodes along the string.
For instance, the third harmonic implies that the string has three antinodes and, accordingly, three nodes including the fixed ends. The presence of these characteristics identifies the pattern of vibration and helps locate points of maximum and zero amplitude (antinodes and nodes).
This is pivotal for calculating displacement, velocity, and acceleration, especially when considering how the string oscillates in stationary waves.
Standing Waves
Standing waves are a phenomenon where the wave appears to be stationary despite being generated by the interference of two traveling waves in opposite directions.
In a string instrument scenario, standing waves form when a vibrating string is confined by fixed ends, meaning waves reflect back upon themselves.
  • Nodes are points on the string with no movement, where destructive interference occurs.
  • Antinodes are points of maximum amplitude, where constructive interference happens.
Understanding the distribution of nodes and antinodes enables us to calculate the amplitude at specific points on the string, which is crucial for determining how the string vibrates and the properties of the sound it produces.
Wave Number
The wave number, often denoted by the symbol \( k \), is an important wave parameter in physics that signifies the number of wave crests in a given distance. It’s defined mathematically as \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength.
This parameter is great for describing how many cycles fit in a unit length of the string. In our exercise, the wave number informs us about the spatial distribution of the standing wave's peaks and troughs on the string.
  • It is essential when using the equation of motion for standing waves \( A(x) = A_0 \sin(kx) \).
  • The wave number has units of radians per meter (rad/m), showing the phase change per unit length.
Calculating \( k \) helps predict the points along the string where the amplitude will be at its peak or zero.
Transverse Velocity
Transverse velocity describes how fast points on a vibrating string move perpendicular to their equilibrium position. In simple terms, it is the speed of oscillation.The maximum transverse velocity \( v_{max} \) can be derived from the formula \( v_{max} = A_0 \cdot \omega \), where \( \omega \) represents the angular frequency (\( \omega = 2\pi f \)).
This calculation is crucial because it indicates the speed at which the string moves vertically as it vibrates.
  • Higher frequency or amplitude increases this velocity.
  • It affects the loudness and pitch of the sound produced by the string.
Understanding transverse velocity is vital for appreciating how energy is transferred through the string as it vibrates.
Transverse Acceleration
Transverse acceleration pertains to how quickly the velocity of a point on the string changes as it moves transversely. It is the acceleration of points along the string perpendicular to their original position.The equation for maximum transverse acceleration is \( a_{max} = A_0 \cdot \omega^2 \), where \( \omega \) is the angular frequency.
This tells us how quickly a point on the string reaches maximum speed in its oscillation cycle.
  • It is particularly high at antinodes, where the displacement is greatest.
  • Like velocity, higher amplitude and frequency contribute to greater acceleration.
Therefore, understanding the concept of transverse acceleration provides insights into how dynamic the movement of the wave is and its potential impact on the structural integrity of the string.

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Most popular questions from this chapter

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?

CALC Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F .\) You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t)=\) 2.30 \(\mathrm{mm} \cos [(6.98 \mathrm{rad} / \mathrm{m}) x+(742 \mathrm{rad} / \mathrm{s}) t] .\) Being more practical, you measure the rope to have a length of 1.35 \(\mathrm{m}\) and a mass of 0.00338 kg. You are then asked to determine the following: (a) amplitude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) average power transmitted by the wave.

A water wave traveling in a straight line on a lake is described by the equation $$y(x, t)=(3.75 \mathrm{cm}) \cos \left(0.450 \mathrm{cm}^{-1} x+5.40 \mathrm{s}^{-1} t\right)$$ where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattem to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

mass 0.0280 \(\mathrm{kg} .\) You measure that it takes 0.0600 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^{7} \mathrm{m} .\) You suspend a lead weight from the lower end of a light string that is 4.00 \(\mathrm{m}\) long and has mass 0.0280 kg. You measure that it takes 0.0600 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that its effect on the tension in the string can be neglected. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

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