Sound — Set 5

Correct Answer: D. 20 Hz to 20 kHz

• **20 Hz to 20 kHz** = the range of sound frequencies detectable by a healthy young human ear; below 20 Hz is infrasound, above 20 kHz is ultrasound. • **Age-related decline**: by age 60, the upper limit often falls to ~12 kHz; presbycusis (age-related hearing loss) is common. • 💡 Wrong-option analysis: 0 Hz to 5 Hz: this is infrasound — far below the audible range; 2 Hz to 200 Hz: this covers only the extreme low end; 200 kHz to 2 MHz: this is ultrasound used in medical imaging — far above audible range.

Correct Answer: A. Greater than 20 kHz

• **Greater than 20 kHz** = ultrasound is sound with frequency above the upper threshold of human hearing (~20 kHz). • **Medical ultrasound** uses 1–20 MHz; industrial applications range from 20 kHz to several MHz depending on penetration needs. • 💡 Wrong-option analysis: Between 20 Hz and 20 kHz only: this is the audible range, not ultrasound; Less than 20 Hz: this is infrasound — the opposite extreme; Exactly 20 Hz: the lower limit of hearing — not ultrasound.

Correct Answer: D. Less than 20 Hz

• **Less than 20 Hz** = infrasound is sound with frequency below the lower threshold of human hearing (~20 Hz). • **Natural sources**: earthquakes, volcanic eruptions, and large atmospheric explosions produce infrasound that can travel thousands of kilometres. • 💡 Wrong-option analysis: Exactly 1 kHz: this is a mid-audible frequency — not infrasound; Greater than 20 kHz: this is ultrasound, not infrasound; Between 20 Hz and 20 kHz only: this is the audible range.

Correct Answer: D. 343 m/s

• **343 m/s** = the speed of sound in dry air at ~20°C; it depends on temperature via v ≈ 331 + 0.6T m/s (T in °C). • **At 0°C**, speed = 331 m/s; the ~0.6 m/s per °C increase means a 10°C rise adds ~6 m/s to the speed. • 💡 Wrong-option analysis: 34.3 m/s: 10× too slow — approximately walking speed; 3.43 m/s: 100× too slow — very slow even for a human; 3430 m/s: 10× too fast — close to the speed in steel, not air.

Correct Answer: D. 1500 m/s

• **1500 m/s** = sound travels much faster in water than in air because water is far less compressible (higher bulk modulus) and denser — but the elasticity effect dominates. • **Water vs. air**: water's bulk modulus ≈ 2.2 GPa vs. air ≈ 142 kPa — a ratio of ~15,000, explaining the ~4× speed advantage. • 💡 Wrong-option analysis: 15000 m/s: 10× too high — close to speed in diamond, not water; 15 m/s: 100× too low — roughly a bicycle speed; 150 m/s: 10× too low — speed in water is clearly in the 1400–1600 m/s range.

Correct Answer: B. 5000 m/s

• **5000 m/s** = sound speed in steel ≈ 5000–6000 m/s because steel's Young's modulus (~200 GPa) and density (~7800 kg/m³) give v = √(E/ρ) ≈ 5100 m/s. • **Speed hierarchy**: steel (~5000 m/s) > water (~1500 m/s) > air (~343 m/s) — solid > liquid > gas. • 💡 Wrong-option analysis: 500 m/s: about 10× too slow — closer to speed in some soft rubbers; 50 m/s: 100× too slow — absurdly low for any solid; 1500 m/s: this is the speed of sound in water, not steel.

Correct Answer: D. Sound Navigation and Ranging

• **Sound Navigation and Ranging** = SONAR is an acronym; it uses transmitted and reflected sound pulses to detect objects and measure distances underwater. • **Active SONAR** emits pulses and listens for echoes; **passive SONAR** only listens for sounds emitted by targets (e.g., submarines). • 💡 Wrong-option analysis: Sound Notation and Recording: invented term — not the SONAR acronym; Signal Navigation and Ranging: wrong first word — it is 'Sound', not 'Signal'; Sound Navigation and Reflection: wrong last word — 'Ranging' is correct, not 'Reflection'.

Correct Answer: B. v = fλ

• **v = fλ** = the fundamental wave equation: wave speed equals the product of frequency and wavelength; this applies to all types of waves. • **Example**: at 1 kHz in air (v = 343 m/s), λ = 343/1000 = 0.343 m — a useful reference wavelength. • 💡 Wrong-option analysis: v = f + λ: dimensionally inconsistent — you cannot add m/s and m; v = f/λ: gives units of s⁻¹/m = Hz/m — wrong; v = λ/f: gives m × s = m·s — wrong units.

Correct Answer: D. 0.1 s

• **0.1 s** = the human ear can distinguish two sounds as separate if the gap between them is ≥ ~0.1 s; at 343 m/s this requires a reflector at least ~17 m away. • **Echo vs. reverberation**: if the delay < 0.1 s, the reflected sound merges with the original and is perceived as reverberation, not a distinct echo. • 💡 Wrong-option analysis: 10 s: vastly too long — a 10-second delay would need a reflector ~1700 m away, uncommon in everyday settings; 0.001 s: too short — the ear fuses sounds within 10 ms; 1 s exactly: far too long for a typical echo scenario.

Correct Answer: C. L = 10 log10(I/I0)

• **L = 10 log₁₀(I/I₀)** = the sound intensity level in decibels; the factor of 10 is the Bel-to-decibel conversion, and I₀ = 10⁻¹² W/m². • **Scale reference**: 0 dB = threshold of hearing; 60 dB = normal conversation; 120 dB = pain threshold; each 10 dB = 10× intensity. • 💡 Wrong-option analysis: L = I/I₀: just a ratio — no logarithm — gives a dimensionless number, not decibels; L = log₁₀(I/I₀): missing the factor 10 — gives Bels, not decibels; L = 20 log₁₀(I₀/I): wrong ratio (inverted) and factor 20 is for amplitude ratios, not intensity ratios.