Sound — Set 6

Correct Answer: D. 1 × 10^-12 W/m^2

• **1 × 10⁻¹² W/m²** = the reference intensity I₀ corresponds approximately to the threshold of hearing for a 1 kHz tone in a young, healthy ear. • **At 0 dB**: I = I₀ = 10⁻¹² W/m²; pain threshold ≈ 1 W/m² = 10¹² × I₀ → 120 dB. • 💡 Wrong-option analysis: 0 W/m²: zero intensity would make the logarithm undefined (log 0 = −∞) — physically meaningless as a reference; 1 W/m²: this is the pain threshold (~120 dB), not the zero-dB reference; 1 × 10⁻⁶ W/m²: this would set 0 dB at 60 dB above the actual hearing threshold — not the standard definition.

Correct Answer: C. 10 times

• **10 times** = from L = 10 log₁₀(I/I₀), a 10 dB increase means log₁₀(I₂/I₁) = 1, so I₂/I₁ = 10. • **Each +10 dB = ×10 in intensity**: 20 dB → ×100; 30 dB → ×1000; 60 dB → ×10⁶. • 💡 Wrong-option analysis: 2 times: this corresponds to ~3 dB increase (10 log₁₀ 2 ≈ 3); 0.1 times: a 10 dB decrease — not increase; 100 times: this corresponds to 20 dB increase, not 10 dB.

Correct Answer: D. |f1 - f2|

• **|f₁ − f₂|** = beats arise from the superposition of two slightly different frequencies; the beat frequency equals the absolute difference, giving the number of loudness maxima per second. • **Practical limit**: beats are distinct when |f₁ − f₂| ≤ ~15 Hz; larger differences are perceived as two separate tones. • 💡 Wrong-option analysis: f₁ × f₂: product of frequencies — has no acoustic meaning as beat frequency; f₁ / f₂: ratio — used for interval identification (e.g., octave = 2), not beats; f₁ + f₂: sum frequency — appears in non-linear mixing but not in simple beat production.

Correct Answer: A. f = v/2L

• **f = v/2L** = in an open organ pipe, the fundamental mode has antinodes at both open ends with one node at the centre, fitting exactly half a wavelength (λ = 2L) in the pipe length L. • **Harmonics**: all integer multiples (f, 2f, 3f, …) are supported in an open pipe — both odd and even. • 💡 Wrong-option analysis: f = v/4L: this is the fundamental of a closed pipe (one closed end) — not an open pipe; f = v/L: this would mean λ = L, i.e., only a half-wave fits — only possible for specific harmonics; f = 2v/L: this would mean λ = L/2 — the second harmonic, not the fundamental.

Correct Answer: C. f = v/4L

• **f = v/4L** = in a closed pipe, the fundamental mode has a node at the closed end and an antinode at the open end, fitting a quarter wavelength (λ = 4L) in the pipe length L. • **Result**: a closed pipe of the same length as an open pipe sounds one octave lower (half the frequency). • 💡 Wrong-option analysis: f = v/L: wavelength would equal L — not the boundary condition for a closed pipe; f = v/2L: this is the fundamental of an open pipe — twice as high in pitch; f = 2v/L: gives λ = L/2 — an overtone, not the fundamental.

Correct Answer: D. Only odd harmonics

• **Only odd harmonics** = a closed pipe (node at closed end, antinode at open end) supports only modes where λ = 4L/n for n = 1, 3, 5, … — only odd integers — so only odd harmonics are present. • **Missing even harmonics** makes the closed pipe sound more hollow or 'woody' compared to an open pipe which supports all harmonics. • 💡 Wrong-option analysis: Only even harmonics: closed pipes suppress even harmonics, not odd ones; All harmonics: only open pipes support all harmonics; No harmonics at all: all real pipes support harmonics above the fundamental.

Correct Answer: A. Speed of object divided by speed of sound

• **Speed of object divided by speed of sound** = Mach number M = v_object / v_sound; M < 1 is subsonic, M = 1 is transonic/sonic, M > 1 is supersonic, M > 5 is hypersonic. • **At Mach 1**, the object is travelling at exactly the speed of sound — pressure waves pile up into a shock wave front. • 💡 Wrong-option analysis: Frequency divided by wavelength: this gives wave speed (v = fλ) — not Mach number; Wavelength divided by frequency: this also gives speed but without the sonic reference; Speed of sound divided by speed of object: this is the inverse of Mach number.

Correct Answer: C. Tympanic membrane

• **Tympanic membrane** = the eardrum separates the outer ear canal from the middle ear; it vibrates in response to sound pressure variations and transmits those vibrations to the ossicles. • **Diameter** ~9 mm; it is a thin, cone-shaped membrane sensitive to pressure changes as small as a few micropascals (threshold of hearing). • 💡 Wrong-option analysis: Cochlea: the spiral fluid-filled organ in the inner ear that converts mechanical vibrations to nerve impulses — not the eardrum; Optic nerve: carries visual signals from the retina to the brain — nothing to do with hearing; Retina: the photosensitive layer of the eye — not part of the auditory system.

Correct Answer: D. About 2–4 kHz

• **About 2–4 kHz** = the human auditory system is most sensitive in the 2–4 kHz range because the ear canal resonance (fundamental at ~3.4 kHz) and the middle ear mechanics both peak here. • **Equal-loudness curves** show that a 1 kHz tone at 40 dB sounds equally loud as a 4 kHz tone at ~35 dB — less intensity is needed near the sensitivity peak. • 💡 Wrong-option analysis: About 0.2–0.4 Hz: this is well below infrasound — inaudible entirely; About 20–40 MHz: this is radiofrequency electromagnetic radiation — far beyond any acoustic range; About 200–400 kHz: this is ultrasound used in sonar/medicine — not audible.

Correct Answer: D. I ∝ 1/r^2

• **I ∝ 1/r²** = for a point source radiating in all directions, power P spreads over a sphere of area 4πr²; intensity I = P/(4πr²) — an inverse-square law. • **Halving distance** quadruples intensity (+6 dB); doubling distance quarters it (−6 dB). • 💡 Wrong-option analysis: I ∝ r²: intensity increasing with distance violates energy conservation for a source with fixed power output; I ∝ 1/r: this would be true only for a line source (e.g., long straight pipe) — not a point source; I ∝ r: intensity increasing with distance — physically impossible for a free-space point source.