Waves Basics — Set 4

Correct Answer: A. Resonance tube

• **Resonance tube** = a tube with an adjustable water level creates air columns of varying length; resonance occurs when the column length matches λ/4 (closed pipe) or multiples — the length at resonance is used to calculate speed. • **Formula**: v = 4L₁f for the first resonance of a closed tube — L₁ is the column length at first resonance. • 💡 Wrong-option analysis: Spherometer: measures radius of curvature of spherical surfaces — not sound; Micrometer screw gauge: measures small lengths (~0.01 mm precision) — not resonance; Vernier calipers: measures external/internal dimensions — not air column resonance.

Correct Answer: D. 20 Hz to 20 kHz

• **20 Hz to 20 kHz** = the typical audible frequency range for a young, healthy human ear; below 20 Hz is infrasound, above 20 kHz is ultrasound. • **Range factor**: 20 kHz / 20 Hz = 1000 — the human ear spans approximately 10 octaves of sound. • 💡 Wrong-option analysis: 20 kHz to 200 kHz: this is entirely in the ultrasound range — not audible; 1 Hz to 10 Hz: this is infrasound — inaudible; 200 Hz to 2 MHz: starts mid-audible range but extends far into ultrasound.

Correct Answer: A. Amplitude

• **Amplitude** = wave energy E ∝ A² for many simple waves (e.g., E = ½kA² for a spring, I = ½ρvω²A² for sound); doubling amplitude quadruples energy. • **Practical implication**: a seismic wave with double the amplitude carries 4× the energy — explains why small increases in magnitude cause large damage increases. • 💡 Wrong-option analysis: Wavelength: energy is not proportional to wavelength²; Frequency: energy per photon = hf (quantum context) but for classical wave energy, A² is the key factor; Time period: inverse of frequency — not directly related to wave energy in classical terms.

Correct Answer: C. Destructive interference

• **Destructive interference** = when two waves of equal amplitude meet in exactly opposite phase (path difference = λ/2, nλ/2 for odd n), the displacements cancel at that point. • **Result**: at points of complete destructive interference, amplitude = 0 — the energy redistributes to constructive-interference regions, not destroyed. • 💡 Wrong-option analysis: Always increased intensity: opposite phase gives decreased, not increased, intensity; A single wave of double speed: wave speed is a medium property — superposition does not change speed; No change at all points: superposition clearly changes displacement at the overlap region.

Correct Answer: A. Propagation of wavefronts

• **Propagation of wavefronts** = Huygens' principle states that every point on a wavefront acts as a source of secondary spherical (or circular) wavelets; the new wavefront is the envelope of these wavelets. • **Applications**: explains reflection (equal angles), refraction (Snell's law), and diffraction — all from one geometric construction. • 💡 Wrong-option analysis: Heat transfer: governed by Fourier's law of conduction — no wavefront concept; Nuclear decay: governed by quantum mechanics and strong/weak nuclear forces; Electric current: governed by Ohm's law and circuit theory.

Correct Answer: B. 50 Hz

• **50 Hz** = f = 1/T = 1/0.02 = 50 Hz; time period 0.02 s means 50 complete oscillations per second. • **Mains frequency**: 50 Hz is the frequency of AC electricity supply in India/Europe — a useful reference. • 💡 Wrong-option analysis: 20 Hz: this would require T = 1/20 = 0.05 s — not 0.02 s; 2 Hz: T = 0.5 s — 25 times too slow; 5 Hz: T = 0.2 s — 10 times too slow.

Correct Answer: B. Ultrasonic waves

• **Ultrasonic waves** = SONAR uses sound frequencies above 20 kHz (ultrasound) because shorter wavelengths give better resolution and reflect well from submerged objects. • **Depth formula**: depth = (v × t) / 2, where v ≈ 1500 m/s in seawater and t = round-trip echo time. • 💡 Wrong-option analysis: Infrared waves: electromagnetic — absorbed strongly by water; not useful for underwater detection; Radio waves: electromagnetic — do not propagate well in seawater; Gamma rays: high-energy ionizing radiation — not used for navigation.

Correct Answer: C. v = fλ

• **v = fλ** = the universal wave equation relating wave speed, frequency, and wavelength; each second, f cycles pass, and each cycle spans λ metres, so v = fλ m/s. • **Valid for**: sound, light, water waves, seismic waves — any sinusoidal wave. • 💡 Wrong-option analysis: v = f + λ: dimensionally inconsistent — cannot add m/s (f) to m (λ); v = f/λ: gives Hz/m — wrong; v = λ/f²: gives m/Hz² — wrong units.

Correct Answer: C. λ/2

• **λ/2** = for a string fixed at both ends, both ends must be nodes (zero displacement); the fundamental mode fits exactly one half-wavelength (one loop) between the two nodes at the fixed ends. • **Fundamental frequency**: f₁ = v/2L; harmonics: fₙ = nv/2L for n = 1, 2, 3, … • 💡 Wrong-option analysis: 2λ: would require L = 2λ — the string would contain 4 half-loops, giving the 4th harmonic; λ: would give f = v/L — the 2nd harmonic, not the fundamental; λ/4: this applies to the fundamental of a pipe closed at one end — not a string fixed at both ends.

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 length L. • **Harmonics**: only odd multiples — f, 3f, 5f, … (n = 1, 3, 5, …) — even harmonics are absent. • 💡 Wrong-option analysis: f = v/L: λ = L — this would require a half-open-half-closed boundary condition not present; f = 2v/L: λ = L/2 — corresponds to the 4th harmonic of an open pipe; f = v/2L: this is the fundamental of an open pipe (both ends open) — twice the closed-pipe frequency.