The problems are on the stationary waves worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A stationary wave forms when two identical waves travel in opposite directions and superpose, giving fixed nodes and antinodes. These worked answers correct the 20 problems of the Physics 11-12 worksheet on stationary waves, from progressive or stationary? to a pipe filled with helium: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the formula, the substitution and the result with unit in bold. Use v = 343 m/s for sound in air unless the problem says otherwise, and remember that the pipe type decides which formula applies.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.4
Problem 1 Progressive or stationary?
1. A progressive wave has the same amplitude at every point and carries energy along the medium. A stationary wave has zero amplitude at nodes, maximum at antinodes, and transfers no net energy along the medium.2. A node is a point of zero amplitude; an antinode is a point of maximum amplitude.
3. Adjacent nodes are λ/2 apart.
Problem 2 Loops on a rope
1. L = 3 × λ/2, so λ = 2L/3 = 2 × 3.0 ÷ 3 = 2 m.2. λ/2 = 1.0 m.
3. 4 nodes (2 ends + 2 in the middle).
Problem 3 Guitar string harmonics
1. f₁ = v/2L = 260 ÷ (2 × 0.65) = 200 Hz.2. f₂ = 2f₁ = 400 Hz.
3. f₃ = 3f₁ = 600 Hz.
Problem 4 Bamboo flute
1. An antinode at each end.2. f₁ = v/2L = 343 ÷ 1.20 = 286 Hz.
3. f₂ = 572 Hz and f₃ = 858 Hz (2f₁ and 3f₁).
Problem 5 Closed organ pipe
1. f₁ = v/4L = 343 ÷ 1.00 = 343 Hz.2. 3f₁ = 1029 Hz and 5f₁ = 1715 Hz.
3. A closed pipe needs a node at the closed end and an antinode at the open end, so an odd number of quarter wavelengths must fit; 2f₁ would put an antinode at both ends or a node at both ends.
Problem 6 Reading a string diagram
1. 3 antinodes, so this is the 3rd harmonic.2. L = 3λ/2, so λ = 2L/3 = 2 × 1.20 ÷ 3 = 0.80 m.
3. v = fλ = 150 × 0.80 = 120 m/s.
Problem 7 Chocolate in the microwave
1. Reflected microwaves form a stationary wave; hot spots are at the antinodes, where the amplitude is largest.2. Antinodes are λ/2 apart, so λ = 2 × 6.1 = 12.2 cm.
3. c = fλ = 2.45 × 10⁹ × 0.122 = 298900000.000 m/s (about 3.0 × 10⁸ m/s).
Problem 8 Sitar string
1. 2.0 g/m = 0.0020 kg/m.2. v = √(T/μ) = √(80 ÷ 0.0020) = √40 000 = 200 m/s.
3. f₁ = v/2L = 200 ÷ 1.60 = 125 Hz.
Problem 9 Tuning the string
1. f₁ = (1/2L)√(T/μ), and L and μ are constant, so f₁ is proportional to √T.2. T ∝ f², so T = 80 × (150 ÷ 125)² = 80 × 1.44 = 115.2 N.
3. (115.2 − 80) ÷ 80 × 100 = 44 %.
Problem 10 Melde string
1. L = 4 × λ/2, so λ = 2 × 1.20 ÷ 4 = 0.6 m.2. v = fλ = 50 × 0.60 = 30 m/s.
3. T = mg = 0.50 × 9.81 = 4.905 N. μ = T/v² = 4.905 ÷ 900 = 0.00545 kg/m (5.4 g/m).
Problem 11 Vocal tract
1. f₁ = v/4L = 343 ÷ 0.68 = 504 Hz; f₂ = 3f₁ = 1513 Hz; f₃ = 5f₁ = 2522 Hz.2. One end (the glottis) is almost closed and acts as a node; the lips are open and form an antinode.
3. Only odd multiples of f₁ appear.
Problem 12 Which pipe?
1. The closed pipe (bottom) has a node at its closed end.2. Open: f₁ = 343 ÷ (2 × 0.50) = 343 Hz. Closed: f₁ = 343 ÷ (4 × 0.50) = 171.5 Hz.
3. 343 ÷ 171.5 = 2, the open pipe is an octave higher.
Problem 13 End correction
1. f₁ = 343 ÷ (4 × 0.30) = 285.8 Hz.2. Leff = 0.30 + 0.6 × 0.020 = 0.312 m.
3. f₁ = 343 ÷ (4 × 0.312) = 274.8 Hz, slightly lower.
Problem 14 Warm air in a pipe
1. v = 331 + 12 = 343 m/s, f₁ = v/2L = 343 ÷ 1.00 = 343 Hz.2. v = 331 + 21 = 352 m/s, f₁ = 352 Hz.
3. (352 − 343) ÷ 343 × 100 = 2.6 %.
Problem 15 Sound against a wall
1. λ = v/f = 343 ÷ 1715 = 0.2 m.2. λ/2 = 0.1 m (10 cm).
3. A node is at the wall, so the nearest antinode is λ/4 = 0.05 m (5.0 cm) away.
Problem 16 Consecutive resonances
1. For a string fixed at both ends the harmonics are f₁, 2f₁, 3f₁… so neighbouring harmonics differ by f₁: 300 − 240 = 60 Hz.2. 240 ÷ 60 = 4 and 300 ÷ 60 = 5.
3. v = 2Lf₁ = 2 × 1.50 × 60 = 180 m/s.
Problem 17 Which type of pipe?
1. 425 ÷ 255 = 5/3, which matches harmonics 3 and 5. In an open pipe the 4th harmonic (340 Hz) would lie between the 3rd and 5th, but nothing resonates between 255 Hz and 425 Hz, so even harmonics are missing: the pipe is closed at one end.2. f₁ = 255 ÷ 3 = 85 Hz (and 425 ÷ 5 = 85 Hz).
3. L = v/4f₁ = 343 ÷ (4 × 85) = 1.01 m.
Problem 18 Frequency against 1/L
1. f₁ = v/2L = (v/2)(1/L) with constant v, which has the form y = mx.2. Gradient = (400 − 160) ÷ (2.50 − 1.00) = 160 Hz·m.
3. gradient = v/2, so v = 320 m/s. T = μv² = 0.0010 × 102400.0000 = 102.4 N.
Problem 19 Spot the error
1. The student used the open-pipe formula for a closed pipe, and then doubled the frequency, but a closed pipe has only odd harmonics.2. f₁ = v/4L = 343 ÷ (4 × 0.80) = 107 Hz.
3. The next resonance is 3f₁ = 322 Hz.
Problem 20 A pipe filled with helium
1. f₁ = v/4L = 343 ÷ 1.36 = 252.2 Hz.2. f₁ = 970 ÷ 1.36 = 713.2 Hz.
3. The vocal tract length stays the same, but the speed of sound is higher in helium, so every resonant frequency rises in the ratio vHe/vair = 2.83.
