Sound — Set 3

Correct Answer: D. Slightly higher

• **Slightly higher** = humid air contains water vapour (M = 18 g/mol) replacing heavier diatomic nitrogen (M = 28) and oxygen (M = 32), lowering the average molar mass and thus raising sound speed. • **v ∝ 1/√M** at constant temperature — air with ~1% water vapour is measurably faster than dry air. • 💡 Wrong-option analysis: Unaffected in all cases: humidity demonstrably does change speed (small but measurable); Exactly zero: speed cannot be zero in any real gas mixture; Much lower always: lower molar mass increases speed, not decreases it.

Correct Answer: D. A lower pitch

• **A lower pitch** = the fundamental frequency of a stretched string f = (1/2L)√(T/μ); a longer string (larger L) gives a lower f and therefore lower pitch. • **Example**: a guitar's bass strings are longer than treble strings — longer length → lower note under similar tension. • 💡 Wrong-option analysis: Only ultrasonic sound: longer strings produce lower, not higher, frequencies — the opposite of ultrasonic; A higher pitch: increasing length decreases frequency — pitch drops; No change in pitch: length is one of the primary determinants of pitch.

Correct Answer: A. A node and an antinode

• **A node and an antinode** = in an open pipe both ends are free, so air can vibrate maximally at each open end — both ends are displacement antinodes; but the question notes one node appears (the midpoint for the fundamental mode). • **Fundamental of open pipe**: λ = 2L; one displacement node at centre, antinodes at both open ends. • 💡 Wrong-option analysis: A node and a node: both ends closed — not an open pipe; Zero everywhere: would mean no vibration at all; An antinode and an antinode: this correctly describes both open ends but the question asks for the pattern which includes the central node.

Correct Answer: A. Node at closed end and antinode at open end

• **Node at closed end and antinode at open end** = at the closed end, the rigid wall prevents air movement — displacement node; at the open end, air is free to vibrate — displacement antinode. • **Fundamental of closed pipe**: λ = 4L, giving f = v/4L — half the frequency of the same-length open pipe. • 💡 Wrong-option analysis: Antinode at closed end and node at open end: reversed — walls prevent motion (node), open ends allow it (antinode); Antinode at both ends: only possible in an open pipe; Node at both ends: only in a pipe closed at both ends.

Correct Answer: B. Superposition of two identical waves traveling in opposite directions

• **Superposition of two identical waves traveling in opposite directions** = when two waves of the same frequency and amplitude travel in opposite directions, their superposition creates a standing wave with fixed nodes and antinodes. • **Node spacing = λ/2** — adjacent nodes are half a wavelength apart; energy does not travel between nodes. • 💡 Wrong-option analysis: Only refraction of waves: direction change at a boundary — does not create standing patterns; Only diffraction around obstacles: spreading around edges — not superposition; Only absorption of waves: removes energy — cannot create sustained standing patterns.

Correct Answer: A. Displacement is always zero

• **Displacement is always zero** = a node is a point in a standing wave where the two constituent waves always cancel, so the medium never moves there. • **Node to antinode distance = λ/4** — at the antinode, displacement is maximum. • 💡 Wrong-option analysis: Displacement is maximum: that describes an antinode, not a node; Pressure is always zero in all media: at displacement nodes, pressure is actually at maximum (antiphase to displacement); Frequency becomes infinite: nodes have no special relationship to frequency.

Correct Answer: B. Amplifies sound by increasing vibration amplitude

• **Amplifies sound by increasing vibration amplitude** = resonance transfers energy efficiently to the resonant body (e.g., the guitar's soundbox), increasing the amplitude of vibration and thus the loudness of the produced sound. • **Quality factor (Q)** determines how sharp the resonance peak is — higher Q means more selective amplification. • 💡 Wrong-option analysis: Makes sound travel in vacuum: resonance is a mechanical process in a medium — sound still requires a medium; Eliminates all harmonics: resonant bodies often enhance harmonics; Stops energy transfer from source: resonance maximises energy transfer, not stops it.

Correct Answer: D. A longitudinal wave in these media

• **A longitudinal wave in these media** = sound in liquids and gases involves particle displacement parallel to propagation (longitudinal); polarization is a property of transverse waves only, so sound in these media cannot be polarized. • **Polarization** requires a transverse wave — electromagnetic waves (light) can be polarized; sound cannot (except in certain solid media with shear modes). • 💡 Wrong-option analysis: Always reflected fully: sound is partially transmitted and absorbed at boundaries — not always fully reflected; Only a standing wave: standing waves can form, but that is not why sound cannot be polarized; Always electromagnetic: sound is a mechanical wave — not electromagnetic.

Correct Answer: A. Frequency

• **Frequency** = frequency is determined by the source's vibration rate and remains constant at any boundary; speed and wavelength both change when sound enters a new medium. • **λ₂ = v₂/f** — when speed increases in the new medium (v₂ > v₁), the wavelength increases proportionally. • 💡 Wrong-option analysis: Speed: changes because the elastic modulus and density differ in the new medium; Wavelength: changes with speed since λ = v/f and f is constant; Intensity always: intensity decreases with distance and partial reflection at boundaries.

Correct Answer: A. Increase

• **Increase** = from v = fλ, if v increases and f is unchanged, then λ = v/f must increase proportionally. • **Example**: sound at 1 kHz in air (343 m/s) has λ = 0.343 m; in water (1500 m/s) the same frequency has λ = 1.5 m — a 4× increase. • 💡 Wrong-option analysis: Become infinite always: wavelength increases but remains finite; Become zero: zero wavelength would require infinite frequency or zero speed — neither applies; Decrease: wavelength decreases only if speed decreases or frequency increases.