Waves, Sound, and Light

26 questions

All types26 Multiple choice13 Matching6 Ordering7

A naval research vessel uses active sonar to determine the depth of the ocean floor. The ship's transducer transmits an acoustic pulse downward into seawater where sound travels at an average speed of $1{}500\text{ m/s}$ . If the reflected echo is recorded by the receiver exactly $4.0\text{ seconds}$ after transmission, what is the depth of the ocean floor?

6{}000\text{ meters}

750\text{ meters}

3{}000\text{ meters}

375\text{ meters}

Show answer & explanation

Answer: $3{}000\text{ meters}$

Answer

3{}000\text{ meters}

Sonar relies on reflected sound waves (echoes). Because the sound pulse must travel down to the seafloor and then back up to the ship's receiver, the

4.0\text{ seconds}

represents two-way travel time. The one-way travel time is

2.0\text{ seconds}

. Multiplying the speed of sound in seawater (

1{}500\text{ m/s}

) by the one-way time (

2.0\text{ s}

) gives an ocean depth of

3{}000\text{ meters}

Step-by-Step Solution

Calculate the total distance traveled by the sound wave during the round trip

\text{Total distance} = \text{speed} \times \text{total time} = 1{}500\text{ m/s} \times 4.0\text{ s} = 6{}000\text{ meters}

The acoustic pulse travels from the ship down to the ocean floor and reflects back up to the surface.

Divide the total round-trip distance by two to obtain the one-way depth

\text{Depth} = \frac{6{}000\text{ m}}{2} = 3{}000\text{ meters}

Sonar distance measurement involves two equal legs of travel (downward and upward).

Key Concept

Calculation of distance using sound wave reflection and two-way travel time

Estimated Time:1m 0s

Question 22Question →

A monochromatic beam of light travels through air and enters a transparent block of crown glass. As the light wave enters the optically denser glass medium, what happens to its speed, wavelength, and frequency?

Its speed increases and its frequency decreases, while its wavelength remains constant.

Its speed decreases and its frequency decreases, while its wavelength remains constant.

Its speed decreases and its wavelength decreases, while its frequency remains constant.

Its speed increases and its wavelength increases, while its frequency remains constant.

Show answer & explanation

Answer: Its speed decreases and its wavelength decreases, while its frequency remains constant.

Answer

Its speed decreases and its wavelength decreases, while its frequency remains constant.

When light transitions from air into a medium with a higher refractive index such as glass, its propagation speed decreases (

v = c/n

). The frequency of the oscillation is an intrinsic property established by the wave source and remains invariant at the interface. By the wave relationship

v = f\lambda

, a reduction in wave velocity at constant frequency results in a proportional reduction in wavelength.

Step-by-Step Solution

Determine the effect of the medium change on wave speed.

The speed of light decreases from

c \approx 3.0 \times 10^8\text{ m/s}

in air to approximately

2.0 \times 10^8\text{ m/s}

in crown glass (

n \approx 1.52

) based on

v = \frac{c}{n}

An increase in optical density (higher index of refraction) reduces the propagation speed of electromagnetic waves.

Determine the behavior of wave frequency across the medium boundary.

The frequency (

f

) remains completely unchanged.

Wave frequency is set solely by the emitting source and is conserved across media boundaries because the number of wave cycles arriving per second must equal the number transmitted.

Apply the wave equation to determine the resulting change in wavelength.

The wavelength (

\lambda

) decreases proportionally with speed according to

\lambda = \frac{v}{f}

With speed

v

reduced and frequency

f

constant, the spatial distance between consecutive crests must shorten.

Key Concept

Wave speed and wavelength changes during medium transition under constant frequency (

v = f\lambda

Estimated Time:45s

Question 23Question →

Electromagnetic waves span a wide range of frequencies, each exhibiting distinct physical properties and practical applications across technology and medicine. Match each type of electromagnetic radiation with its primary characteristic or practical use.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Infrared Radiation matches thermal energy and thermal imaging; Ultraviolet Radiation matches vitamin D synthesis and sunburns; X-rays match skeletal medical imaging through soft tissue penetration; Radio Waves match broadcast transmissions and long-wavelength telecommunications.

Each type of electromagnetic radiation is characterized by distinct wavelength, frequency, and photon energy bands. Radio waves have the longest wavelengths and lowest energy, suitable for broadcasting; infrared radiation transfers thermal heat; ultraviolet radiation induces photochemical reactions in skin; and X-rays have sufficient penetration power to create shadow radiographs through soft tissue.

Step-by-Step Solution

Identify the characteristic behavior and application of infrared radiation.

Infrared radiation is associated with molecular thermal vibration and heat emission, used in night-vision and thermal sensors.

Infrared photons are emitted by warm bodies and absorbed by matter as sensible heat.

Identify the biological effects and energy level of ultraviolet radiation.

Ultraviolet radiation carries higher photon energy than visible light, stimulating vitamin D synthesis in the skin but causing cellular damage leading to sunburns.

UV photon energy is sufficient to initiate chemical bond changes in biological tissue.

Identify the high-energy penetration characteristics of X-rays.

X-rays pass easily through soft tissues like muscle and skin but are blocked by dense bone structure.

High-frequency electromagnetic waves have high penetration power that is stopped predominantly by high-density matter.

Identify the fundamental wave traits of radio waves.

Radio waves possess the lowest frequencies and longest wavelengths, making them ideal for long-distance broadcast communication.

Long wavelengths diffract readily around terrain and experience low atmospheric absorption.

Key Concept

Electromagnetic Spectrum Characteristics and Applications

Estimated Time:1m 0s

Question 24Question →

An observer at a construction site watches a worker strike a steel pipe with a hammer and hears the sound transmitted through the pipe before hearing it through the open air. Which of the following best explains why sound travels faster through solid steel than through air?

The frequency of the sound wave increases significantly when traveling through denser matter.

The closely packed particles and high elasticity of solids allow vibrational energy to transfer more rapidly between molecules.

Sound waves convert into electromagnetic transverse waves when moving through solid materials.

The amplitude of the wave increases automatically because solid matter absorbs less acoustic energy.

Show answer & explanation

Answer: The closely packed particles and high elasticity of solids allow vibrational energy to transfer more rapidly between molecules.

Answer

Sound travels faster through solid steel than through air because the closely packed particles and high elasticity of solids allow mechanical vibrational energy to transfer more rapidly between molecules.

Sound is a mechanical wave that propagates through the physical displacement and elastic restoration of medium particles. In solids like steel, atoms are tightly bonded in a crystal lattice with high elastic modulus, enabling compressions and rarefactions to transfer rapidly from particle to particle. In contrast, gas molecules in air are widely spaced and require longer travel times between collisions, resulting in a much lower propagation speed.

Step-by-Step Solution

Identify the nature of sound waves and the medium requirement

Sound is a mechanical wave that propagates via sequential collisions and oscillations of particles in a physical medium.

Understanding that mechanical waves rely on intermolecular interactions is necessary to evaluate the effect of the medium on wave speed.

Compare particle arrangement and elasticity between solids and gases

Molecules in steel are closely bound with high elastic modulus, whereas air molecules are widely separated with low elasticity.

Greater medium elasticity and density of packing directly increase the speed of acoustic transmission (

v = \sqrt{E/\rho}

Relate medium properties to wave speed differences

Sound travels at roughly

5{}960\text{ m/s}

in steel compared to approximately

343\text{ m/s}

in air.

The rapid transfer of kinetic energy between adjacent atoms in the solid explains why the sound arrives first through the steel pipe.

Key Concept

Mechanical sound wave propagation speed depends on the elasticity and atomic density of the transmitting medium, traveling fastest in solids, slower in liquids, and slowest in gases.

Estimated Time:1m 0s

Question 25Question →

The visible light spectrum consists of electromagnetic radiation spanning various wavelengths and frequencies. Arrange the following colors of visible light in order of increasing frequency, from lowest frequency to highest frequency.

Drag items to arrange them in the correct order

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Answer

The correct order from lowest frequency to highest frequency is Red light, Yellow light, Blue light, and Violet light.

Electromagnetic waves in the visible spectrum have frequencies that are inversely related to their wavelengths. Red light has the longest wavelength (around

700\text{ nm}

) and thus the lowest frequency. As wavelength decreases across the spectrum toward violet (around

400\text{ nm}

), wave frequency increases progressively from yellow to blue to violet.

Step-by-Step Solution

Recall the fundamental wave relationship connecting wave speed, frequency, and wavelength.

Since all electromagnetic waves travel at the same speed in a vacuum (

c = f\lambda

), frequency is inversely proportional to wavelength (

f = c / \lambda

This relationship demonstrates that colors with the longest wavelengths have the lowest frequencies, while colors with the shortest wavelengths have the highest frequencies.

Determine the relative wavelengths and frequencies of the given visible light colors using the standard spectral sequence.

From longest to shortest wavelength, the visible spectrum follows the sequence: Red

\rightarrow

Orange

\rightarrow

Yellow

\rightarrow

Green

\rightarrow

Blue

\rightarrow

Indigo

\rightarrow

Violet.

Mapping the colors to this sequence directly determines their relative positions regarding frequency.

Arrange the specified colors in order of increasing frequency.

The sequence from lowest frequency (longest wavelength) to highest frequency (shortest wavelength) is Red light

\rightarrow

Yellow light

\rightarrow

Blue light

\rightarrow

Violet light.

This orders the items accurately according to the specified physical criteria.

Key Concept

Visible Light Spectrum and Wave Frequency

Estimated Time:1m 0s

Question 26Question →

A sound wave generated in air enters a pool of water, where its propagation speed increases by a factor of four. How are the frequency and wavelength of the sound wave affected upon entering the water?

The frequency quadruples, while the wavelength remains unchanged.

The frequency remains unchanged, while the wavelength quadruples.

The frequency remains unchanged, while the wavelength is reduced to one-fourth.

Both the frequency and the wavelength double.

Show answer & explanation

Answer: The frequency remains unchanged, while the wavelength quadruples.

Answer

The frequency remains unchanged, while the wavelength quadruples.

When any wave crosses the boundary between two media, its frequency is conserved because frequency is determined solely by the source vibration. According to the wave equation

v = f\lambda

, if wave speed increases by a factor of 4 while frequency remains constant, the wavelength must also increase by a factor of 4 to maintain the equality.

Step-by-Step Solution

Identify the behavior of wave frequency across a boundary.

The frequency (

f

) remains constant because it depends exclusively on the oscillating source.

When a wave transitions from one medium into another, particles at the boundary oscillate in synchronization with the incident wave, preserving frequency.

Apply the universal wave equation relating speed, frequency, and wavelength.

v = f\lambda

, which rearranges to

\lambda = \frac{v}{f}

This formula defines the relationship between the physical parameters of any wave.

Calculate the proportional change in wavelength when speed increases by a factor of 4.

Since

v_{\text{water}} = 4v_{\text{air}}

and

f

is constant,

\lambda_{\text{water}} = \frac{4v_{\text{air}}}{f} = 4\lambda_{\text{air}}

Wavelength is directly proportional to wave speed when frequency is constant.

Key Concept

Wave boundary behavior and the wave equation (

v = f\lambda

)

Estimated Time:1m 0s

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Waves, Sound, and Light

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