This question tests the independence of wave parameters in linear media. In a linear medium, wave speed depends only on medium properties (like density and elasticity), not on wave amplitude. Since both waves travel through the same fluid at the same frequency, they have identical wave speeds and wavelengths. The key difference is energy: wave energy is proportional to amplitude squared, so Wave 2 carries more energy while maintaining the same propagation characteristics. Choice A incorrectly links amplitude to wave speed, confusing energy content with propagation speed. Remember: in linear media, changing amplitude affects energy and intensity but not frequency, wavelength, or wave speed.

This question tests the effect of damping on oscillatory motion. Damping represents energy dissipation mechanisms that convert mechanical energy into other forms (typically heat), reducing the system's total mechanical energy over time. Greater damping means energy is lost more rapidly, resulting in smaller oscillation amplitudes for the same initial conditions. The natural frequency may slightly decrease with heavy damping, but this doesn't increase amplitude. Choice B incorrectly claims damping adds energy to the system, contradicting the fundamental nature of dissipative forces. Remember: damping always removes energy from oscillatory systems, reducing amplitude over time—it never adds energy or increases amplitude.

This question tests standing wave patterns for strings fixed at both ends. The second harmonic (n=2) has a wavelength λ = L, where L is the string length, creating two half-wavelengths along the string. This produces two antinodes (maximum displacement points) with a node at the center, plus the required nodes at both fixed ends. The fundamental (n=1) would have only one antinode, while the third harmonic (n=3) would have three antinodes. Choice D describes the third harmonic pattern, not the second. A useful check: for the nth harmonic on a string fixed at both ends, there are n antinodes and (n+1) nodes total.

This question tests the Doppler effect for sound waves when the source (blood) moves toward the observer (probe). When a source moves toward an observer, the wavefronts are compressed, decreasing the effective wavelength between successive crests. Since wave speed in the medium remains constant (v = 1540 m/s in tissue), and v = fλ, a decrease in wavelength must correspond to an increase in frequency. The received frequency is therefore higher than the emitted frequency f₀, confirming choice B. Choice A incorrectly states that motion toward the observer increases wavelength, which would occur if the source moved away. Choice C ignores the Doppler effect entirely. Choice D incorrectly links frequency shift to amplitude, when the Doppler shift depends only on relative motion. For Doppler problems, remember: source moving toward observer → higher frequency; source moving away → lower frequency. The shift magnitude depends on the ratio of source speed to wave speed.

This question tests wave interference based on path difference. When two coherent sources emit in phase, the phase difference at the observation point depends on the path difference. A path difference of ΔL = λ/2 corresponds to a phase difference of π radians (180°), since phase difference = (2π/λ) × path difference = (2π/λ) × (λ/2) = π. When two waves of equal amplitude arrive with opposite phase (π phase difference), they interfere destructively, producing zero net amplitude at that point. This confirms choice B is correct. Choice A incorrectly claims λ/2 path difference produces constructive interference, which would require integer multiples of λ. Choice C incorrectly denies that sound waves can interfere. Choice D incorrectly links path length to amplitude rather than phase. For interference problems, remember: path difference of nλ → constructive; path difference of (n + 1/2)λ → destructive, where n is an integer.

This question tests mechanical wave properties, specifically the relationship between frequency, wavelength, and speed. The wave speed in a medium is constant and determined by the medium's properties, with v = fλ holding for sinusoidal waves. In this fluid-filled tube modeling airway mucus, doubling the frequency while keeping the medium unchanged maintains constant wave speed. The correct answer B follows because halving the wavelength satisfies v = fλ when f doubles and v is fixed. Distractor A fails due to the misconception that wavelength increases with frequency, but they are inversely related at constant speed. To verify in similar questions, calculate wavelength changes using λ = v/f. Always confirm if the medium's properties are altered, as that affects v.

This question tests mechanical wave propagation and resolution in ultrasound. Higher frequency reduces wavelength via λ = v/f, improving resolution as smaller wavelengths distinguish closer features. In this ultrasound phantom, increasing frequency with constant v shortens λ for better spatial resolution. The correct answer B follows because shorter wavelength directly enhances resolution without altering speed or reflection. Distractor A assumes speed increases with frequency, a misconception ignoring that v depends on the medium. For other wave resolution problems, recall resolution improves with shorter λ. Check if frequency changes affect attenuation, though not relevant here.

This question tests wave interference principles. Destructive interference for equal-amplitude coherent waves occurs when path difference δ = (m + 1/2)λ, a half-integer multiple. In this audiology setup, the condition for cancellation matches this. The correct answer B follows because destructive interference requires odd multiples of λ/2 path difference. Distractor A describes constructive interference, a common confusion of conditions. For interference problems, use δ = mλ for constructive, (m+1/2)λ for destructive. Assume coherence unless stated otherwise.

This question tests resonance in wave systems like the cochlea. Maximal displacement occurs where local natural frequency matches stimulus frequency. Increasing stimulus frequency shifts peak to higher-frequency regions. The correct answer A follows because resonance mapping places higher f at corresponding sites. Distractor B ignores frequency dependence, assuming fixed location. In tonotopic models, map frequency to position via resonance. Check if system has graded properties affecting local frequencies.

This question tests harmonics in standing waves. For fixed ends, nth harmonic has n antinodes, λ=2L/n; three antinodes is n=3, λ=2L/3. With three antinodes, it's the third harmonic. The correct answer A follows because it matches the harmonic and wavelength. Distractor B miscounts as second with wrong λ, confusing node count. In standing wave questions, count antinodes for n. Use f_n = n v /(2L) for frequency scaling.