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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 \(0.0750 \mathrm{~s}\). 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) The adjacent nodes are \(15.0 \, \mathrm{cm}\) apart. b) The wavelength is \(30.0 \, \mathrm{cm}\) or \(0.3 \, \mathrm{m}\), the amplitude is \(0.850 \, \mathrm{cm}\) or \(0.00850 \, \mathrm{m}\), and the speed is \(4.00 \, \mathrm{m/s}\). c) The maximum and minimum transverse speeds are \(0.71 \, \mathrm{m/s}\) and \(0 \, \mathrm{m/s}\) respectively. d) The shortest distance between a node and an antinode is \(7.5 \, \mathrm{cm}\).

Step by step solution

01

Find the distance between adjacent nodes

The distance between successive antinodes corresponds to wavelength/2. Therefore, the distance between successive nodes is also equal to wavelength/2. Hence, the distance between the adjacent nodes is \(15.0 \, \mathrm{cm}\), the same as the antinode distance.
02

wavelength, amplitude, and speed of traveling wave

As previously determined, the distance between successive antinodes corresponds to wavelength/2. So the wavelength of the wave (\(\lambda\)) is \(2 \times 15.0 \, \mathrm{cm} = 30.0 \, \mathrm{cm} = 0.3 \, \mathrm{m}\). The amplitude of the wave (A) is given as \(0.850 \, \mathrm{cm} = 0.00850 \, \mathrm{m}\). The speed of a wave (v) can be found using the formula \(v = \lambda / T\), where T is the period. Substituting given values will yield the speed to be \(0.3 \, \mathrm{m} / 0.0750 \, \mathrm{s} = 4.00 \, \mathrm{m/s}\).
03

maximum and minimum transverse speeds

The speed of a particle undergoing simple harmonic motion is given by \(v = A \omega \cos(\omega t)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency and \(t\) is the time. The maximum speed is achieved when \(cos(\omega t) = 1\), therefore, \(v_{\mathrm{max}} = A \omega\). The angular frequency \(\omega\) can be found from the period \(T\) of oscillation using the formula \(\omega = 2\pi / T\). Substituting all values gives \(v_{\mathrm{max}} = 0.00850 \, \mathrm{m} \times (2\pi / 0.0750 \, \mathrm{s}) = 0.71 \, \mathrm{m/s}\). The minimum speed at the antinode is 0, as the particle momentarily stops at the peak and trough of the oscillation.
04

shortest distance between a node and an antinode

The shortest distance along the string between a node and an antinode is half the distance from one node to the next, or half the distance from one antinode to the next. Therefore, it equals to half the answer from step 1, which is \(15.0 \, \mathrm{cm} / 2 = 7.5 \, \mathrm{cm}\).

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Key Concepts

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

Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of oscillatory motion that occurs when the restoring force acting on a particle is directly proportional to the displacement from its equilibrium position and acts in the direction opposite to the displacement. In the context of a standing wave, at the antinodes, particles undergo simple harmonic motion as they move back and forth. This type of motion can be characterized by properties such as amplitude, period, and frequency.
The amplitude in SHM is the maximum displacement of the particle from its equilibrium position. The period is the time it takes for the particle to complete one full oscillation. Frequency, which is the inverse of the period, represents how many oscillations occur in one second. The angular frequency is related to these properties and is given by \[ \omega = \frac{2\pi}{T} \]where \(T\) is the period.

In our problem, the period of oscillation is given as 0.0750 s, which allows us to compute the angular frequency. This understanding of simple harmonic motion is crucial for determining the behavior of particles at antinodes in a wave.
Wavelength
Wavelength is the physical length of one complete cycle of a wave. It is usually denoted by the Greek letter lambda, \( \lambda \). Wavelength is a crucial concept in wave mechanics as it helps in understanding the spatial properties of waves. For a standing wave, the distance between successive antinodes or nodes is half of the wavelength.This means if the distance between antinodes is given as 15.0 cm, the complete wavelength can be calculated as\[ \lambda = 2 \times 15.0 \, \text{cm} = 30.0 \, \text{cm} = 0.3 \, \text{m} \]

Wavelength is essential in calculating other wave properties like wave speed. When learning about waves, always remember that wavelength is the distance over which the wave shape repeats.
Amplitude
The amplitude of a wave is the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In simpler terms, it's how "tall" the wave is from the center line. Amplitude is important as it is related to the energy carried by the wave; higher amplitude means more energy.
In the problem given, the amplitude at an antinode is 0.850 cm, which converts to 0.00850 m. In the context of a standing wave, this amplitude is useful for determining the maximum speed at which particles move.
  • The amplitude conveys information about the wave's power and visibility.
  • It's signified with the letter \(A\) and affects the simple harmonic motion's maximum speed, calculated as: \[ v_{\text{max}} = A \omega \]
Understanding amplitude allows us to find how aggressively a wave vibrates and thus helps in applications ranging from sound to light.
Wave Speed
Wave speed is how fast a wave propagates through a medium. Mathematically, wave speed \(v\) is calculated by\[ v = \lambda \times f \] where \( f \) is the frequency of the wave. Alternatively, you can use \[ v = \frac{\lambda}{T} \] where \( T \) is the wave period.
The exercise provides enough information to find the wave speed. Given \(\lambda = 0.3 \, \text{m}\) and \(T = 0.0750 \, \text{s}\), we find the speed to be \[ v = \frac{0.3 \, \text{m}}{0.0750 \, \text{s}} = 4.00 \, \text{m/s} \].

Understanding wave speed is crucial because it tells us how quickly information or energy travels. Whether in sound, light, or water waves, calculating wave speed allows us to predict wave interactions and understand their behavior across different media.

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Most popular questions from this chapter

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.

Standing waves are produced on a string that is held fixed at both ends. The tension in the string is kept constant. (a) For the second overtone standing wave the node-to-node distance is \(8.00 \mathrm{~cm} .\) What is the length of the string? (b) What is the node-to-node distance for the fourth harmonic standing wave?

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is \(0.026 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(4.3 \mathrm{~m}\) from the source. (a) What is the intensity at a distance of \(3.1 \mathrm{~m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

A horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's mass?

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes \(2.5 \mathrm{~s}\) for the boat to travel from its highest point to its lowest, a total distance of \(0.53 \mathrm{~m} .\) The fisherman sees that the wave crests are spaced \(4.8 \mathrm{~m}\) 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 \(0.30 \mathrm{~m}\) but the other data remained the same, how would the answers to parts (a) and (b) change?

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