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Chapter 16: Problem 3

A longitudinal wave with a frequency of \(3.0 \mathrm{~Hz}\) takes \(1.7 \mathrm{~s}\) to travel the length of a \(2.5-\mathrm{m}\) Slinky (see Figure \(16-3\) ). Determine the wavelength of the wave.

Short Answer

Expert verified
The wavelength is approximately 0.49 m.

Step by step solution

01

Identify Given Variables

We have the frequency of the wave as \( f = 3.0 \, \mathrm{Hz} \) and the time taken to travel the slinky is \( t = 1.7 \, \mathrm{s} \). The length of the slinky is \( L = 2.5 \, \mathrm{m} \).
02

Calculate Wave Speed

The speed \( v \) of the wave can be calculated using the formula \( v = \frac{L}{t} \). Substitute in the given length and time: \( v = \frac{2.5 \, \mathrm{m}}{1.7 \, \mathrm{s}} \approx 1.47 \, \mathrm{m/s} \).
03

Use Wave Speed to Find Wavelength

The relationship between wave speed \( v \), frequency \( f \), and wavelength \( \lambda \) is given by the formula \( v = f \cdot \lambda \). Solve for the wavelength: \( \lambda = \frac{v}{f} \).
04

Substitute Values into Wavelength Equation

Substitute the wave speed and frequency into the formula: \( \lambda = \frac{1.47 \, \mathrm{m/s}}{3.0 \, \mathrm{Hz}} \approx 0.49 \, \mathrm{m} \).
05

Solution Conclusion

The wavelength of the wave traveling through the slinky is approximately \( 0.49 \, \mathrm{m} \).

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Key Concepts

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

Longitudinal Waves
A longitudinal wave is a type of wave where the particles of the medium move parallel to the direction of wave propagation. This is different from transverse waves, where the particles move perpendicular to the wave's direction. Longitudinal waves are common in environments such as sound waves traveling through air or waves in a slinky.

These waves compress and rarefy the medium they move through, creating areas of high and low pressure. Imagine a slinky stretched out on a table. If you push one end, the coils bunch up and then spread out as the wave travels. This movement is characteristic of longitudinal waves.

Key characteristics of longitudinal waves include:
  • **Compression and rarefaction**: Wave regions where particles are close together and then spread apart.
  • **Medium movement**: Particles oscillate parallel to the direction of the wave.
  • **Examples**: Sound waves, seismic P-waves, and waves in springs or slinkies.
Wave Speed Calculation
Wave speed is how fast the wave propagates through a medium. It is an integral concept in understanding how waves move. The speed of a wave can be calculated using the formula \( v = \frac{L}{t} \), where \( v \) is the wave speed, \( L \) is the distance traveled by the wave, and \( t \) is the time taken to cover that distance.

In the slinky example, the wave travels a length of 2.5 meters in 1.7 seconds. Plugging these values into the formula gives a wave speed of approximately 1.47 meters per second. Calculating wave speed helps in determining other wave properties such as wavelength and frequency, integral to understanding the wave's behavior in different media.

Key aspects to consider while calculating wave speed include:
  • **Distance covered**: Accurate measurement of the path length the wave has traveled.
  • **Time taken**: Precise timing of how long it takes for the wave front to pass.
  • **Formula application**: Utilizing \( v = \frac{L}{t} \) correctly ensures accurate results.
Wavelength Determination
Determining the wavelength of a wave is crucial for comprehensively understanding its behavior and impact. Wavelength, denoted as \( \lambda \), is the spatial period of the wave鈥攖he distance over which the wave's shape repeats.

In the context of longitudinal waves traveling through a slinky, once we know the wave speed and frequency, we use the formula \( v = f \cdot \lambda \) to find the wavelength. Rearranging gives \( \lambda = \frac{v}{f} \). Substituting our previously calculated wave speed of 1.47 m/s and frequency of 3.0 Hz, we find the wavelength to be approximately 0.49 meters.

Some points to keep in mind while determining wavelength are:
  • **Wave speed**: Always verify the speed is accurately calculated.
  • **Frequency**: Ensure the frequency is correct and in Hertz (Hz).
  • **Correct application of the formula**: Use \( \lambda = \frac{v}{f} \) properly to solve for the wavelength.

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Most popular questions from this chapter

Suppose that sound is emitted uniformly in all directions by a public address system. The intensity at a location \(22 \mathrm{~m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). What is the intensity at a spot that is \(78 \mathrm{~m}\) away?

co Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) \(\mathrm{A}\) small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y\), and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope Be sure that your answers are consistent with your answer to Concept Question (b).

The wavelength of a sound wave in air is \(2.74 \mathrm{~m}\) at \(20{ }^{\circ} \mathrm{C}\). What is the wavelength of this sound wave in fresh water at \(20{ }^{\circ} \mathrm{C} ?\) (Hint: The frequency of the sound is the same in both media.)

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{~m}^{2},\) has a linear density of \(7.0 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of \(46 \mathrm{~m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). What will be the speed of the wave when the temperature is lowered by \(14 \mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 440 -Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of \(2.0 \mathrm{~s}\). The difference between the maximum and minimum sound frequencies detected by the microphone is \(2.1 \mathrm{~Hz}\). Ignoring any reflections of sound in the room and using \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound, determine the amplitude of the simple harmonic motion.
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