Waves, Sound, and Light Flashcards

Two waves with a wavelength of 0.40 m travel through the same medium and meet at point P. Wave A travels 1.80 m to reach point P, and Wave B travels 1.00 m to reach point P. What type of interference occurs at point P?

Answer: Constructive interference, because the path difference equals exactly 2 wavelengths

The path difference is 1.80 m − 1.00 m = 0.80 m. Dividing by the wavelength: 0.80 m ÷ 0.40 m = 2. When the path difference is an integer number of wavelengths (0, 1, 2, …), the waves arrive perfectly in phase and undergo fully constructive interference. A non-integer multiple (0.5, 1.5, 2.5, …) would produce destructive interference.

A fire truck moving at 30 m/s toward a stationary observer sounds a siren at 800 Hz. The speed of sound in air is 340 m/s. Which expression correctly calculates the frequency the observer hears?

Answer: 800 × 340 / (340 − 30)

When the source moves toward a stationary observer, the Doppler formula is f_obs = f_s × v / (v − v_s). The denominator uses subtraction because the moving source compresses the wavefronts ahead of it, shortening the wavelength and raising the observed frequency. Answer A applies to a moving observer (not a moving source), and Answer D gives the frequency for a source moving away.

A pipe that is open at both ends has a length of 0.85 m. If the speed of sound is 340 m/s, what is the frequency of the third harmonic produced by this pipe?

Answer: 600 Hz

For a pipe open at both ends, both ends are antinodes, so the allowed harmonics follow f_n = nv / (2L), where n = 1, 2, 3, … For the third harmonic: f₃ = 3 × 340 / (2 × 0.85) = 1,020 / 1.70 = 600 Hz. A common mistake is using the closed-pipe formula f_n = nv / (4L), which applies only to pipes closed at one end.

A tuning fork of unknown frequency is struck alongside a standard 256 Hz fork, producing 4 beats per second. When a small piece of wax is applied to the unknown fork (lowering its frequency slightly), the beat frequency drops to 2 beats per second. What was the original frequency of the unknown fork?

Answer: 260 Hz

The unknown fork is either 252 Hz or 260 Hz (both 4 Hz away from 256 Hz). Adding wax lowers the unknown fork's frequency. If the unknown were 252 Hz, lowering it further would increase the gap from 256 Hz and the beats would increase — but they decreased. Therefore the unknown must be above 256 Hz. At 260 Hz, lowering it slightly (say to ~258 Hz) brings it closer to 256 Hz, reducing beats to 2 Hz. The original frequency is 260 Hz.

Light travels from Medium A (index of refraction n = 1.6) into Medium B (index of refraction n = 1.2). What is the critical angle for total internal reflection at this interface?

Answer: 48.6°

The critical angle θ_c satisfies sin(θ_c) = n₂ / n₁ = 1.2 / 1.6 = 0.75, so θ_c = arcsin(0.75) ≈ 48.6°. Total internal reflection only occurs when light travels from a denser medium to a less dense medium (n₁ > n₂) and the angle of incidence exceeds this critical angle. The distractors correspond to common errors such as inverting the ratio (sin⁻¹(1.6/1.2) is undefined) or using n values for common glass (1.5) instead of the given values.

A photon has a wavelength of 600 nm in vacuum. It then enters a glass medium with an index of refraction of 1.5. Which statement correctly describes what changes and what stays the same inside the glass?

Answer: Speed and wavelength both decrease; frequency stays the same

When light enters a medium, its speed decreases to v = c/n = c/1.5 ≈ 2×10⁸ m/s. Since frequency is set by the source and must be conserved across the boundary (to maintain continuity of the wave), frequency does not change. Because v = fλ, a lower speed with constant frequency forces the wavelength to decrease: λ_glass = 600 nm / 1.5 = 400 nm. A common misconception is that frequency changes; it is always the wavelength that adjusts.