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Chapter 15: Problem 70

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 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
Calculate amplitude and use given equations to find velocity and acceleration at specified points.

Step by step solution

01

Understand the Harmonic Pattern

Since the string is vibrating in its third harmonic, there are three antinodes and two nodes between the fixed ends of the string. This is important because the amplitude varies sinusoidally between nodes and antinodes.
02

Amplitude Calculation - General Equation

The amplitude at any point on the string can be found using the equation: \[ A(x) = A_0 \sin\left(\frac{n\pi x}{L}\right) \]where \(A_0\) is the amplitude at the antinode (0.400 cm), \(n\) is the harmonic number (3 for the third harmonic), \(x\) is the distance from the left end, and \(L\) is the length of the string.
03

Calculate Amplitude at Specific Points

(i) For \(x = 40.0\) cm:\[ A(40.0) = 0.400 \sin\left(\frac{3\pi \times 40.0}{\text{L}}\right) \](ii) For \(x = 20.0\) cm:\[ A(20.0) = 0.400 \sin\left(\frac{3\pi \times 20.0}{\text{L}}\right) \](iii) For \(x = 10.0\) cm:\[ A(10.0) = 0.400 \sin\left(\frac{3\pi \times 10.0}{\text{L}}\right) \]Without the length \(L\), we use the harmonic information to solve these equations.
04

Calculate Full Cycle Time

The time for the string to go from the largest upward displacement to the largest downward displacement is half the period of the wave. The period \(T\) can be calculated by \( T = \frac{1}{f} \), where \(f\) is the frequency (240 Hz). \[ T = \frac{1}{240} \, \text{s} \]Half of this is the time sought: \( \frac{1}{2 \times 240} \, \text{s} \).
05

Calculate Maximum Transverse Velocity

The maximum transverse velocity \(v_{max}\) at any point can be calculated using:\[ v_{max} = 2\pi f A(x) \]Substitute \(A(x)\) for each of the calculated amplitudes from Step 3 to find the maximum velocity at each point.
06

Calculate Maximum Transverse Acceleration

The maximum transverse acceleration \(a_{max}\) can be found using:\[ a_{max} = \left(2\pi f\right)^2 A(x) \]Calculate this value using each \(A(x)\) from Step 3 for the points (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm.

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Key Concepts

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

Harmonic Motion
Harmonic motion refers to the repetitive oscillation that occurs in systems like springs and waves when displaced from an equilibrium position. In the context of the vibrating string, it vibrates in its third harmonic. This means that the string forms a standing wave with three antinodes and two nodes. A standing wave arises when waves of the same frequency and amplitude travel through a medium but in opposite directions. These waves overlap and interfere, creating fixed points called nodes (where there is no movement) and points of maximum movement called antinodes.

For a string fixed at both ends, the number of antinodes corresponds to the harmonic number, making the third harmonic have three antinodes. The wave鈥檚 frequency determines how fast this motion repeats, and the string's behavior allows for calculations of other properties like amplitude at various points along the string. Understanding harmonic motion is key in observing how energy is distributed along the string and how wave characteristics behave in different harmonic modes.
Transverse Velocity
The transverse velocity of a point on a wave describes the speed at which the point moves perpendicular to the direction of the wave's travel. In a standing wave, this is particularly important, as every point on the string oscillates between positive and negative directions as the waves interfere.

For our vibrating string, transverse velocity reaches its maximum where the motion is most vigorous, at the antinodes. This maximum transverse velocity is calculated using the formula:
  • \[ v_{max} = 2 \, \pi \, f \, A(x) \]
Here, \(f\) is the given frequency of the wave, and \(A(x)\) is the amplitude at the specific point on the string. As the wave progresses through one full cycle, the amplitude and frequency work together to determine how fast a point on the string reaches its maximum speed. Understanding the maximum transverse velocity aids in determining the dynamic behavior of the wave at different points.
Transverse Acceleration
Transverse acceleration is the rate of change of transverse velocity for a point on the wave. It represents how quickly the speed of a point changes as it moves up and down along the string. In the third harmonic of a vibrating string, this acceleration varies across the string's length.

The maximum transverse acceleration at any point is found by squaring the factor of frequency in the transverse velocity formula, giving us the equation:
  • \[ a_{max} = (2 \, \pi \, f)^2 \, A(x) \]
Here the constants ensure the acceleration scales with both frequency and amplitude. The frequency \(f\), which is 240 Hz in this scenario, significantly influences the acceleration. Since acceleration occurs during changes in velocity direction, it reaches its maximum at the nodes. High transverse acceleration indicates rapid changes in velocity, an essential feature of energetic oscillations found in standing waves. By grasping the concepts of maximum transverse acceleration, students can appreciate the forces involved in wave motion on a fixed string.

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Most popular questions from this chapter

For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s for the boat to travel from its highest point to its lowest, a total distance of 0.53 m. The fisherman sees that the wave crests are spaced 4.8 m apart. (a) How fast are the waves traveling? (b) What is the amplitude of each wave? (c) If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) change?

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the \(+x\)-axis and is fixed at \(x = 0\). (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (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?
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