Waves, Sound, and Light Flashcards

A sound wave travels from air into water. Which of the following correctly describes what happens to its frequency and wavelength?

Answer: Frequency stays the same; wavelength increases

When a wave crosses a boundary between two media, its frequency is determined by the source and does not change. However, since speed increases in water (≈1480 m/s vs ≈343 m/s in air) and wavelength = speed / frequency, the wavelength increases proportionally with the speed.

Two identical speakers emit a 680 Hz tone in phase. A listener stands at a point where one speaker is exactly 0.5 m farther away than the other. If the speed of sound is 340 m/s, what does the listener hear?

Answer: Silence due to destructive interference

The wavelength of a 680 Hz sound at 340 m/s is λ = 340/680 = 0.5 m. A path difference of 0.5 m equals exactly one wavelength (λ), which would normally give constructive interference — but wait: 0.5 m = 1λ means the waves arrive in phase, producing constructive interference. Actually, re-examining: path difference = 0.5 m = 1λ → constructive. However, if path difference = λ/2 = 0.25 m → destructive. Here path difference = 0.5 m = 1λ, so the answer is constructive interference (louder sound). The correct answer is A.

Two identical speakers emit a 680 Hz tone in phase. A listener stands at a point where one speaker is exactly 0.25 m farther away than the other. If the speed of sound is 340 m/s, what does the listener hear?

Answer: Silence due to destructive interference

The wavelength is λ = v/f = 340/680 = 0.5 m. A path difference of 0.25 m = λ/2. When two in-phase waves arrive with a half-wavelength path difference, they are perfectly out of phase (one crest meets one trough), producing complete destructive interference. The listener hears silence.

A bat emits a 50,000 Hz ultrasonic pulse and detects the echo from a stationary wall 0.034 seconds later. If the speed of sound is 340 m/s, how far away is the wall?

Answer: 5.78 m

The pulse travels to the wall and back, so total distance = speed × time = 340 × 0.034 = 11.56 m. Since this is the round-trip distance, the one-way distance to the wall is 11.56 / 2 = 5.78 m. The frequency is a distractor — it is not needed to solve this problem.

Which phenomenon best explains why the sky appears blue during the day but the sun appears red/orange near the horizon at sunset?

Answer: Rayleigh scattering — shorter wavelengths scatter more, and at sunset light travels through more atmosphere

Rayleigh scattering causes light to scatter off gas molecules proportional to 1/λ⁴, meaning shorter (blue) wavelengths scatter far more than longer (red) wavelengths. During the day, scattered blue light fills the sky. At sunset, sunlight travels through a much longer path of atmosphere, and blue light is scattered away almost entirely before reaching your eyes, leaving the longer red and orange wavelengths dominant.

A guitar string vibrates at its fundamental frequency of 200 Hz. A musician lightly touches the string at its exact midpoint, forcing a node there, and plucks it again. What frequency does it now produce?

Answer: 400 Hz — the second harmonic (first overtone) is produced

By touching the midpoint, the musician forces a node at the center of the string. The longest standing wave that fits this constraint has nodes at both ends and the middle, corresponding to half the string's length per loop — this is the second harmonic (first overtone), which has exactly twice the frequency of the fundamental: 2 × 200 = 400 Hz. This technique is called playing a harmonic or 'flageolet' and is used in string instruments.