Skip to main content
Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 17, Problem 37

A 2.0-m-long string vibrates at its second-harmonic frequency with a maximum amplitude of 2.0 cm. One end of the string is at x = 0 cm. Find the oscillation amplitude at x = 10, 20, 30, 40, and 50 cm.

Verified step by step guidance
1
Step 1: Understand the problem. The string is vibrating in its second harmonic, which means it has two antinodes and one node (excluding the fixed ends). The wave equation for the standing wave can be expressed as y(x, t) = A * sin(kx) * cos(ωt), where A is the maximum amplitude, k is the wave number, and ω is the angular frequency. The amplitude at any point x is determined by the term A * sin(kx).
Step 2: Determine the wave number k. The wave number k is related to the wavelength λ by the equation k = 2π/λ. For the second harmonic, the wavelength is equal to the length of the string (L) divided by 2, so λ = L/2. Substituting L = 2.0 m, we find λ = 1.0 m. Therefore, k = 2π/λ = 2π/1.0 = 2π rad/m.
Step 3: Write the expression for the amplitude at any position x. The amplitude at position x is given by A(x) = A_max * sin(kx), where A_max is the maximum amplitude of the wave (2.0 cm) and k = 2π rad/m. Convert A_max to meters: A_max = 2.0 cm = 0.02 m.
Step 4: Calculate the amplitude at each specified position. Substitute x = 10 cm, 20 cm, 30 cm, 40 cm, and 50 cm into the equation A(x) = A_max * sin(kx). Convert x to meters before substituting: x = 10 cm = 0.10 m, x = 20 cm = 0.20 m, and so on. For each x, calculate sin(kx) using k = 2π rad/m.
Step 5: Interpret the results. The amplitude at each position will vary depending on the sine function. Note that the amplitude will be zero at nodes (where sin(kx) = 0) and maximum at antinodes (where sin(kx) = ±1). Use this understanding to verify the calculated amplitudes at each position.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Harmonic Frequencies

Harmonic frequencies refer to the specific frequencies at which a system, such as a vibrating string, can oscillate. The second harmonic, for instance, is the first overtone and occurs when the string vibrates in two segments, producing a frequency that is twice that of the fundamental frequency. Understanding these harmonics is essential for analyzing the behavior of the string at various points along its length.
Recommended video:
Guided course
05:08
Circumference, Period, and Frequency in UCM

Amplitude of Oscillation

Amplitude is the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In this context, the maximum amplitude of 2.0 cm indicates how far the string moves from its rest position during oscillation. The amplitude can vary along the length of the string, especially in different harmonic modes, affecting the displacement at specific points.
Recommended video:
Guided course
04:24
Amplitude Decay in an LRC Circuit

Wave Equation and Node/Antinode Locations

The wave equation describes how waves propagate through a medium, and in the case of a vibrating string, it helps determine the positions of nodes (points of no displacement) and antinodes (points of maximum displacement). For the second harmonic, there are nodes at both ends of the string and an antinode at the center. This understanding is crucial for calculating the oscillation amplitude at various points along the string.
Recommended video:
Guided course
06:28
Equations for Transverse Standing Waves
Related Practice
Textbook Question

Two strings are adjusted to vibrate at exactly 200 Hz. Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

1616
views
Textbook Question

A flute player hears four beats per second when she compares her note to a 523 Hz tuning fork (the note C). She can match the frequency of the tuning fork by pulling out the 'tuning joint' to lengthen her flute slightly. What was her initial frequency?

1628
views
Textbook Question

Two out-of-phase radio antennas at x=±300 m on the x-axis are emitting 3.0 MHz radio waves. Is the point (x, y) =(300 m, 800 m) a point of maximum constructive interference, maximum destructive interference, or something in between?

274
views
Textbook Question

A violinist places her finger so that the vibrating section of a 1.0 g/m string has a length of 30 cm, then she draws her bow across it. A listener nearby in a 20°C room hears a note with a wavelength of 40 cm. What is the tension in the string?

1307
views
Textbook Question

INT One end of a 75-cm-long, 2.5 g guitar string is attached to a spring. The other end is pulled, which stretches the spring. The guitar string's second harmonic occurs at 550 Hz when the spring has been stretched by 5.0 cm. What is the value of the spring constant?

1866
views
Textbook Question

Tendons are, essentially, elastic cords stretched between two fixed ends. As such, they can support standing waves. A woman has a 20-cm-long Achilles tendon—connecting the heel to a muscle in the calf—with a cross-section area of 90 mm2 . The density of tendon tissue is 1100 kg/m3. For a reasonable tension of 500 N, what will be the fundamental frequency of her Achilles tendon?

1326
views