Examkin

Waves, Sound, and Light

26 soru

Tüm türler26Çoktan seçmeli13Eşleştirme6Sıralama7
Soru 21Soru →

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 1500 m/s1{}500\text{ m/s}. If the reflected echo is recorded by the receiver exactly 4.0 seconds4.0\text{ seconds} after transmission, what is the depth of the ocean floor?

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Cevap: 3000 meters3{}000\text{ meters}

Cevap

3000 meters3{}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 seconds4.0\text{ seconds} represents two-way travel time. The one-way travel time is 2.0 seconds2.0\text{ seconds}. Multiplying the speed of sound in seawater (1500 m/s1{}500\text{ m/s}) by the one-way time (2.0 s2.0\text{ s}) gives an ocean depth of 3000 meters3{}000\text{ meters}.

Adım Adım Çözüm

1
Calculate the total distance traveled by the sound wave during the round trip
Total distance=speed×total time=1500 m/s×4.0 s=6000 meters\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.
2
Divide the total round-trip distance by two to obtain the one-way depth
Depth=6000 m2=3000 meters\text{Depth} = \frac{6{}000\text{ m}}{2} = 3{}000\text{ meters}
Sonar distance measurement involves two equal legs of travel (downward and upward).

Anahtar Kavram

Calculation of distance using sound wave reflection and two-way travel time
Tahmini Süre:1m 0s
Soru 22Soru →

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?

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

Cevap

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/nv = 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=fλv = f\lambda, a reduction in wave velocity at constant frequency results in a proportional reduction in wavelength.

Adım Adım Çözüm

1
Determine the effect of the medium change on wave speed.
The speed of light decreases from c≈3.0×108 m/sc \approx 3.0 \times 10^8\text{ m/s} in air to approximately 2.0×108 m/s2.0 \times 10^8\text{ m/s} in crown glass (n≈1.52n \approx 1.52) based on v=cnv = \frac{c}{n}.
An increase in optical density (higher index of refraction) reduces the propagation speed of electromagnetic waves.
2
Determine the behavior of wave frequency across the medium boundary.
The frequency (ff) 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.
3
Apply the wave equation to determine the resulting change in wavelength.
The wavelength (λ\lambda) decreases proportionally with speed according to λ=vf\lambda = \frac{v}{f}.
With speed vv reduced and frequency ff constant, the spatial distance between consecutive crests must shorten.

Anahtar Kavram

Wave speed and wavelength changes during medium transition under constant frequency (v=fλv = f\lambda).
Tahmini Süre:45s
Soru 23Soru →

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.

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Öğeler

Infrared Radiation
Ultraviolet Radiation
X-rays
Radio Waves

Eşleşmeler

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Cevap

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.

Adım Adım Çözüm

1
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.
2
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.
3
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.
4
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.

Anahtar Kavram

Electromagnetic Spectrum Characteristics and Applications
Tahmini Süre:1m 0s
Soru 24Soru →

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?

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

Cevap

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.

Adım Adım Çözüm

1
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.
2
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=E/ρv = \sqrt{E/\rho}).
3
Relate medium properties to wave speed differences
Sound travels at roughly 5960 m/s5{}960\text{ m/s} in steel compared to approximately 343 m/s343\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.

Anahtar Kavram

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.
Tahmini Süre:1m 0s
Soru 25Soru →

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.

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Cevap

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 nm700\text{ nm}) and thus the lowest frequency. As wavelength decreases across the spectrum toward violet (around 400 nm400\text{ nm}), wave frequency increases progressively from yellow to blue to violet.

Adım Adım Çözüm

1
Recall the fundamental wave relationship connecting wave speed, frequency, and wavelength.
Since all electromagnetic waves travel at the same speed in a vacuum (c=fλc = f\lambda), frequency is inversely proportional to wavelength (f=c/λf = 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.
2
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.
3
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.

Anahtar Kavram

Visible Light Spectrum and Wave Frequency
Tahmini Süre:1m 0s
Soru 26Soru →

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?

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

Cevap

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=fλv = 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.

Adım Adım Çözüm

1
Identify the behavior of wave frequency across a boundary.
The frequency (ff) 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.
2
Apply the universal wave equation relating speed, frequency, and wavelength.
v=fλv = f\lambda, which rearranges to λ=vf\lambda = \frac{v}{f}.
This formula defines the relationship between the physical parameters of any wave.
3
Calculate the proportional change in wavelength when speed increases by a factor of 4.
Since vwater=4vairv_{\text{water}} = 4v_{\text{air}} and ff is constant, λwater=4vairf=4λair\lambda_{\text{water}} = \frac{4v_{\text{air}}}{f} = 4\lambda_{\text{air}}.
Wavelength is directly proportional to wave speed when frequency is constant.

Anahtar Kavram

Wave boundary behavior and the wave equation (v=fλv = f\lambda)
Tahmini Süre:1m 0s
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Waves, Sound, and Light Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 2 | Examkin