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Chapter 15: Problem 36

15.36 .. CALC Adjacent antinodes of a standing wave on a string are 15.0 \(\mathrm{cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 \(\mathrm{cm}\) 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?

Short Answer

Expert verified
(a) 15.0 cm, (b) 30.0 cm, 0.850 cm, 400 cm/s, (c) 71.21 cm/s and 0 cm/s, (d) 7.5 cm.

Step by step solution

01

Understanding the Problem

We need to analyze a standing wave on a string to solve for various parameters, including the distance between nodes, wavelength, amplitude, speed of traveling waves, and various speeds of a point at an antinode.
02

Relationship Between Antinodes and Nodes

Antinodes in a standing wave are located at half the wavelength (\( \frac{\lambda}{2} \)) from each other. Thus, the wavelength \( \lambda \) can be calculated using the given distance between antinodes, 15.0 cm: \[ \lambda = 2 \times 15.0 \text{ cm} = 30.0 \text{ cm} \] The distance between adjacent nodes is half the wavelength: \[ \text{Distance between nodes} = \frac{\lambda}{2} = 15.0 \text{ cm} \]
03

Calculate Wavelength, Amplitude, and Speed of the Waves

Reaffirming from Step 2, the wavelength is 30.0 cm. The amplitude of each traveling wave creating the standing wave is the same as the amplitude of the oscillation of a particle at an antinode, 0.850 cm. The speed \( v \) of the waves can be calculated from the wavelength and period \( T \): \[ v = \frac{\lambda}{T} = \frac{30.0 \text{ cm}}{0.0750 \text{ s}} = 400 \text{ cm/s} \]
04

Calculate Maximum and Minimum Transverse Speeds at an Antinode

The maximum transverse speed \( v_{max} \) of a particle at an antinode is given by: \[ v_{max} = \omega A \] where \( \omega = \frac{2\pi}{T} \) and \( A \) is the amplitude. Using \( \omega \): \[ \omega = \frac{2\pi}{0.0750} \approx 83.78 \, \text{rad/s} \]\[ v_{max} = 83.78 \times 0.850 = 71.21 \text{ cm/s} \]Since the minimum speed of a particle at an antinode is simply when it's at its turning point, it momentarily stops, so \( v_{min} = 0 \text{ cm/s} \).
05

Shortest Distance Between Node and Antinode

The shortest distance along the string between a node and an adjacent antinode is one quarter of the wavelength, since \( \frac{\lambda}{4} \) of a complete wave cycle is from a node to an antinode:\[ \frac{\lambda}{4} = \frac{30.0 \text{ cm}}{4} = 7.5 \text{ cm} \]

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Key Concepts

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

Wavelength Calculation
When dealing with standing waves on a string, understanding how to calculate the wavelength is pivotal. A standing wave is formed by two traveling waves moving in opposite directions at the same speed, and the location where the amplitude is maximized is known as an antinode. In the problem given, the antinodes are 15.0 cm apart. Since antinodes are positioned at half the wavelength (\( \frac{\lambda}{2} \)) apart, you can easily find the wavelength (\( \lambda \)) by doubling the distance between antinodes:
  • Given that the distance between antinodes is 15.0 cm, the wavelength is calculated as:\[ \lambda = 2 \times 15.0 \text{ cm} = 30.0 \text{ cm} \]
The wavelength is an essential measurement, as it describes the length of one complete wave cycle between repeating points, such as from crest to crest.
Simple Harmonic Motion
Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement. This motion is characteristic of the oscillation seen at the antinode in a standing wave. In this exercise, a particle located at an antinode oscillates with an amplitude of 0.850 cm and a period of 0.0750 seconds. These two values are crucial aspects:
  • Amplitude: This is the maximum displacement of the particle from its resting position鈥攈ere, it's 0.850 cm.
  • Period: This is the time taken for one complete cycle of motion鈥攊n this problem, 0.0750 seconds define this cycle.
During this type of motion, particles at the antinode move back and forth with maximum energy. The formula for angular frequency (\( \omega \)) is \( \omega = \frac{2\pi}{T} \) where \( T \) is the period, allowing us to calculate how fast the particles are oscillating in radians per second.
Wave Speed Calculation
Wave speed (\( v \)) indicates how fast a wave travels along a string. For standing waves, the wave speed can be found using the relationship between wavelength and period:
  • The formula used is \[ v = \frac{\lambda}{T} \] where \( \lambda \) is the wavelength and \( T \) is the period.
In this case, the calculated values were:
  • Wavelength: 30.0 cm
  • Period: 0.0750 s
By substituting these into the formula, the wave speed is:
  • \[ v = \frac{30.0 \text{ cm}}{0.0750 \text{ s}} = 400 \text{ cm/s} \]
This speed reflects how quickly each point of the wave (like the crest or trough) moves across the string.
Node and Antinode Distance
Nodes and antinodes are key features of standing waves. A node is a point on the string where there is no movement, while an antinode is where the maximum movement occurs. Knowing the distance between these points helps to understand the wave's behavior. In the given exercise, the shortest distance from a node to an antinode is one quarter of the wavelength:
  • The formula is: \[ \text{Distance} = \frac{\lambda}{4} \]
Thus, for a wavelength of 30.0 cm, the distance is:
  • \[ \frac{30.0 \text{ cm}}{4} = 7.5 \text{ cm} \]
This distance tells us how quickly maximum displacement transitions into no displacement along the string. Understanding this concept provides insight into the spatial structure of standing waves.

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Most popular questions from this chapter

CALC A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=\) \((5.60 \mathrm{cm}) \sin [(0.0340\) rad/ \(\mathrm{cm}) x] \sin [(50.0 \mathrm{rad} / \mathrm{s}) t],\) where the origin is at the left end of the string, the \(x\) -axis is along the string. and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

A deep-sea diver is suspended beneath the surface of Loch Ness by a \(100-\) m-long cable that is attached to a boat on the surface (Fig. P15.84). The diver and his suit have a total mass of 120 \(\mathrm{kg}\) and a volume of 0.0880 \(\mathrm{m}^{3} .\) The cable has a diameter of 2.00 \(\mathrm{cm}\) and a linear mass density of \(\mu=\) 1.10 \(\mathrm{kg} / \mathrm{m} .\) The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) ) exerts on him. (b)Calculate the tension in the cable a distance \(x\) above the diver. The buoyant force on the cable must be included in your calculation. (c) The speed of transverse waves on the cable is given by \(v=\sqrt{F / \mu}\) (Eq. 15.13). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.. Integrate to find the time required for the first signal to reach the surface.

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?

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?

Threshold of Pain. You are investigating the report of a UFO landing in an isolated portion of New Mexico, and you encounter a strange object that is radiating sound waves uniformly in all directions. Assume that the sound comes from a point source and that you can ignore reflections. You are slowly walking toward the source. When you are 7.5 \(\mathrm{m}\) from it, you measure its intensity to be 0.11 \(\mathrm{W} / \mathrm{m}^{2}\) . An intensity of 1.0 \(\mathrm{W} / \mathrm{m}^{2}\) is often used as the "threshold of pain." How much closer to the source can you move before the sound intensity reaches this threshold?
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