/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 A sound wave travels in air towa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does not change when the sound enters the water. The wavelength of the sound is \(2.74 \mathrm{m}\) in the air, and the temperature of both the air and the water is \(20^{\circ} \mathrm{C} .\) What is the wavelength in the water?

Short Answer

Expert verified
The wavelength of the sound wave in water is approximately 11.84 meters.

Step by step solution

01

Understand the Problem

The problem involves a sound wave traveling from air to water. We know the frequency remains constant during this transition, and we need to find the new wavelength in water given the initial wavelength in air.
02

Recall the Speed of Sound Formula

The speed of sound is related to its frequency and wavelength by the formula: \( v = f \lambda \), where \( v \) is the speed, \( f \) is the frequency, and \( \lambda \) is the wavelength.
03

Determine Speed of Sound in Air and Water

Calculate or look up the speed of sound in air (343 m/s) and water (1482 m/s) at \(20^{\circ} C\). These values are important for calculating the new wavelength.
04

Calculate Frequency Using Air Data

Use the formula \( v = f \lambda \) for air to find the frequency: \( f = \frac{v_{\text{air}}}{\lambda_{\text{air}}} = \frac{343}{2.74} \approx 125.18 \text{ Hz} \).
05

Calculate Wavelength in Water

Use the same frequency and the speed of sound in water to find the new wavelength: \( \lambda_{\text{water}} = \frac{v_{\text{water}}}{f} = \frac{1482}{125.18} \approx 11.84 \text{ m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Wavelength
When discussing sound waves, the wavelength is a crucial concept. It describes the distance between consecutive points of a wave, like crests or troughs. Think of it as the length of one complete wave cycle.
In the original problem, the wavelength of the sound in air is given as 2.74 meters. This is the distance over which the wave pattern repeats. When a sound wave travels from air into water, although its speed changes, the frequency remains the same because it's determined by the source of the sound.
- **Wavelength and Speed:** Upon entering a new medium, the speed of the wave changes but not its frequency. As a result, the wavelength adjusts to match the new speed. This relationship is a fundamental aspect of wave behavior and is expressed by the formula: \[\lambda = \frac{v}{f}\]- **Key Insight:** In the exercise, because the speed of sound in water is higher than in air, the wavelength in water becomes longer. This is because the wave covers more distance in the same amount of time due to the increased speed.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium. It is influenced by the medium's properties, such as density and temperature.
- **Sound in Different Media:** In our problem, the sound wave transitions from air to water. In general, sound travels faster in water than in air primarily because water is denser, allowing sound waves to transmit more efficiently. At 20°C, sound travels at approximately 343 meters per second (m/s) in air and 1482 m/s in water.
Understanding this concept is critical when solving problems involving sound traveling through different media. Given the fixed frequency as the sound moves from air to water, the speed of sound's increased value in water leads to a corresponding change in wavelength to maintain the same frequency. This is crucial for waves maintaining a steady beat despite changing speeds.- **Application of Speed Equation:** The equation relating speed to wavelength and frequency, \( v = f\lambda \), is pivotal here. It helps calculate how the wavelength changes by using known speeds and constant frequency, as demonstrated in the step-by-step solution.
Frequency
Frequency is a measure of how many wave cycles pass a point in one second and is measured in hertz (Hz). It represents the sound wave's pitch—higher frequencies correspond to higher pitches.
- **Frequency Consistency:** One critical aspect of the problem is that the frequency does not change as the wave moves from air to water. This constancy is because the frequency is set by the source and remains the same regardless of the medium.- **Calculating Frequency:** In the solution, we use the speed of sound in air and the given wavelength to find the frequency. Using the formula: \[ f = \frac{v}{\lambda} \] Substituting in the values gives us the frequency of around 125.18 Hz in both air and water.
- **Importance of Frequency:** The unchanged frequency means that any alterations in speed result in a change in wavelength. This understanding allows us to calculate new wave properties like the wavelength of sound in water, providing a broader appreciation for how sound behaves across different environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One of the most important concepts we encountered in this chapter is the transverse wave. For instance, transverse waves travel along a guitar string when it is plucked or along a violin string when it is bowed. Problem 112 reviews how the travel speed depends on the properties of the string and on the tension in it. Problem 113 illustrates how the Doppler effect arises when an observer is moving away from or toward a stationary source of sound. In fact, we will see that it's possible for both situations to occur at the same time. A siren, mounted on a tower, emits a sound whose frequency is \(2140 \mathrm{Hz}\). A person is driving a car away from the tower at a speed of \(27.0 \mathrm{m} / \mathrm{s}\). As the figure illustrates, the sound reaches the person by two paths: the sound reflected from the building in front of the car, and the sound coming directly from the siren. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). Concepts: (i) One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or reflected sound, does the wavelength change? (ii) Why does the driver hear a frequency for the reflected sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz} ?\) (iii) Why does the driver hear a frequency for the direct sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz}\) ? Calculations: What frequency does the person hear for the (a) reflected and (b) direct sound?

A water-skier is moving at a speed of \(12.0 \mathrm{m} / \mathrm{s}\). When she skis in the same direction as a traveling wave, she springs upward every \(0.600 \mathrm{s}\) because of the wave crests. When she skis in the direction opposite to the direction in which the wave moves, she springs upward every 0.500 s in response to the crests. The speed of the skier is greater than the speed of the wave. Determine (a) the speed and (b) the wavelength of the wave.

A monatomic ideal gas \((\gamma=1.67)\) is contained within a box whose volume is \(2.5 \mathrm{m}^{3} .\) The pressure of the gas is \(3.5 \times 10^{s} \mathrm{Pa}\). The total mass of the gas is \(2.3 \mathrm{kg}\). Find the speed of sound in the gas.

At what temperature is the speed of sound in helium (ideal gas, \(\gamma=1.67,\) atomic mass \(=4.003\) u) the same as its speed in oxygen at \(0^{\circ} \mathrm{C} ?\)

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is \(1522 \mathrm{m} / \mathrm{s}\). When the dolphin is swimming directly away at \(8.0 \mathrm{m} / \mathrm{s},\) the marine biologist measures the number of clicks occurring per second to be at a frequency of \(2500 \mathrm{Hz}\). What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.