IB Physics C.4: Standing waves and resonance worked solutions

Points inside one loop rise and fall together, so they are in phase. A is false: the amplitude grows from zero at a node to its largest value at the antinode. B is false: a standing wave carries no net energy along the string. D is false: adjacent nodes are half a wavelength apart.

Fourth harmonic: four loops, nodes at 0, 0.5, 1.0, 1.5 and 2.0\,\mathrm{m}. One node (0.5\,\mathrm{m}) lies between P and Q, so they are in antiphase: \pi. Two nodes (0.5 and 1.0\,\mathrm{m}) lie between P and R, so they are in phase: 0. D uses the travelling-wave rule 2\pi d/\lambda, which does not apply to a standing wave.

Corks that stay still sit at nodes; adjacent nodes are \lambda/2 apart, so \lambda = 6.0\,\mathrm{cm} and v = f\lambda = 8.0\times0.060 = 0.48\,\mathrm{m\,s^{-1}}. B takes the gap as a whole wavelength; D takes it as a quarter-wavelength; A as one eighth.

(a) At an antinode the two travelling waves always arrive in phase, so their amplitudes add: 2A = 8.0\,\mathrm{cm}, A = 4.0\,\mathrm{cm} (A1). (b) The straight rope is the moment when the two waves cancel everywhere. A node has zero amplitude and stays at rest all the time; the antinode points are crossing equilibrium at maximum speed and are 8.0\,\mathrm{cm} out again a quarter of a period later (A1).

First harmonic with nodes at both ends: half a wavelength fits, \lambda = 2L = 0.80\,\mathrm{m}, so v = f\lambda = 440\times0.80 = 352\,\mathrm{m\,s^{-1}}. B takes \lambda = L; D takes \lambda = 4L (the unlike-ends rule); A takes \lambda = L/2.

Three antinodes means the third harmonic, so f_1 = 360/3 = 120\,\mathrm{Hz} and the allowed frequencies are 120, 240, 360, 480\,\mathrm{Hz}, …. Only 240\,\mathrm{Hz} is on the list. B comes from treating 360\,\mathrm{Hz} as the second harmonic.

Open ends are antinodes. With alike ends the nth harmonic has n nodes inside the pipe, so this is the third harmonic: L = 3\lambda/2, \lambda = 2L/3 = 0.60\,\mathrm{m}. B (\lambda = L/2) has four nodes; D uses the unlike-ends rule 4L/3.

The closed end must be a node and the open end an antinode, so the tube holds 1, 3, 5, … quarter-wavelengths and only odd harmonics occur: 3f_1 = 330\,\mathrm{Hz} and 5f_1 = 550\,\mathrm{Hz}. A and C include even harmonics.

Damping transfers energy out of the oscillator, so the amplitude falls. The frequency is set by the mass, the spring and the damping, none of which change, so it is the same in every cycle (very slightly below the undamped value).

A steady forced oscillation always has the driving frequency, 1.5\,\mathrm{Hz}. C is the natural frequency (1/0.50); A reads the period as a frequency; D adds the two frequencies.

The drum's rotation drives the panel; at one spin speed the driving frequency matches the panel's natural frequency, and the large amplitude makes noise and wears fixings. A, B and D use resonance on purpose.

(a) The wave reflects at the wall and travels back along the rope (M1). Incident and reflected waves have the same frequency and wavelength and move in opposite directions, so they superpose into a pattern with fixed nodes and antinodes (A1). (b) \lambda = v/f = 6.0/2.5 = 2.4\,\mathrm{m} (M1); adjacent nodes are \lambda/2 = 1.2\,\mathrm{m} apart (A1). (c) Four loops: 4\times1.2 = 4.8\,\mathrm{m} (A1). (d) T = 1/f = 0.40\,\mathrm{s} (M1). At T/4 every point is passing through equilibrium, so the rope is a straight line; at T/2 every loop has flipped, so the shape is the t=0 shape upside down with the nodes unchanged (A1).

(a) Third harmonic: L = 3\lambda/2, so \lambda = 1.2\,\mathrm{m} (M1); v = f\lambda = 150\times1.2 = 180\,\mathrm{m\,s^{-1}} (A1). (b) Nodes at 0, 0.6, 1.2 and 1.8\,\mathrm{m}; one node (0.60\,\mathrm{m}) lies between X and Z, so they are in antiphase: \pi (A1). (c) Y at 0.30\,\mathrm{m} is halfway between the nodes at 0 and 0.60\,\mathrm{m}, so it is an antinode with amplitude 5.0\,\mathrm{mm}; X is only 0.15\,\mathrm{m} from a node, so its amplitude is smaller (A1). (d) 5.0/2 = 2.5\,\mathrm{mm} (A1).

(a) Nodes at both ends: \lambda = 2L = 0.650\,\mathrm{m} (M1); v = f\lambda = 440\times0.650 = 286\,\mathrm{m\,s^{-1}} (A1). (b) Same tension, so the same wave speed; \lambda = 2\times0.260 = 0.520\,\mathrm{m} (M1); f = 286/0.520 = 550\,\mathrm{Hz} (A1). (c) \lambda_{\text{air}} = 340/440 = 0.773\,\mathrm{m} (A1). The sound has the frequency of the string, but it travels at the speed of sound in air, not the speed of waves on the string, so its wavelength differs (A1).

(a) Node at the closed end, antinode at the open end, one quarter-wavelength along the pipe (A1). (b) L = \lambda/4, so \lambda = 2.04\,\mathrm{m} (M1); f_1 = 340/2.04 = 167\,\mathrm{Hz} (A1). (c) 333\,\mathrm{Hz} = 2f_1, an even harmonic (M1). With unlike ends the pipe can hold only an odd number of quarter-wavelengths, so the allowed frequencies are 167, 500, 833\,\mathrm{Hz}, …; the claim is wrong (A1).

(a) \lambda = 340/262 = 1.30\,\mathrm{m} (M1). P has alike ends: L = \lambda/2 = 0.649\,\mathrm{m} (A1). Q has unlike ends: L = \lambda/4 = 0.324\,\mathrm{m} (A1). (b) The length and the ends of P fix \lambda_1 = 2L (M1), so f_1 = v/2L rises in proportion to v: 262\times1.020 = 267\,\mathrm{Hz} (A1).

(a) The fork drives the air column at 426\,\mathrm{Hz} (M1). The column resonates, with a large amplitude, only when a standing wave fits with the water surface as a node and the open top as an antinode, that is, when 426\,\mathrm{Hz} equals one of the column's natural frequencies (A1). (b) Successive resonant lengths differ by \lambda/2 (M1): \lambda = 2(0.600 - 0.200) = 0.800\,\mathrm{m} (A1); v = f\lambda = 426\times0.800 = 341\,\mathrm{m\,s^{-1}} (A1). (c) Next length = 0.600 + 0.400 = 1.00\,\mathrm{m} (M1), which is less than 1.20\,\mathrm{m}: yes, at an air column of 1.00\,\mathrm{m}, or 5\lambda/4 (A1).

(a) Curve starts at a small, non-zero amplitude (M1) and has a single peak at or very near 4.0\,\mathrm{Hz}, falling below its starting value by 8\,\mathrm{Hz} (A1). (b) The peak is lower and wider (M1), with its maximum at a slightly lower frequency; far from 4.0\,\mathrm{Hz} the two curves almost coincide (A1). (c) A steady forced oscillation has the driving frequency: 6.0\,\mathrm{Hz} (A1).

(a) All three curves start at the same displacement with zero gradient (M1). Light damping: oscillates about zero with decreasing amplitude and a constant period (A1). Critical damping reaches zero first without crossing it; heavy damping creeps back more slowly and stays above the critical curve (A1). (b) Critical damping: the reading can be taken in the shortest time and the pointer never swings past it (A1).

(a) One jolt per slab, so the driving frequency is f = v/d (M1). Resonance when v/0.45 = 2.0: v = 0.90\,\mathrm{m\,s^{-1}} (A1). (b) f = 1.6/0.45 = 3.6\,\mathrm{Hz} (M1); this is far from the natural frequency, so the jolts are not in step with the body's motion, little energy is added per cycle and the amplitude stays small (A1). (c) Fit damping, such as foam pads between body and axle, which dissipate energy every cycle; or stiffen the springs to move the natural frequency away from 2.0\,\mathrm{Hz} (A1).

(a) Alike ends give f_n = nf_1, so neighbouring resonances differ by f_1 = 70\,\mathrm{Hz} (M1). Then 105\,\mathrm{Hz} would be a whole multiple of 70\,\mathrm{Hz}, but 105/70 = 1.5; so the ends differ (A1). (b) Unlike ends allow only odd harmonics, so neighbours differ by 2f_1: f_1 = 70/2 = 35\,\mathrm{Hz} (M1); check: 105 = 3\times35 (A1). L = \lambda_1/4 = v/4f_1 = 343/140 = 2.45\,\mathrm{m} (A1). (c) 245/35 = 7: the seventh harmonic (A1).

(a) From largest displacement to straight is a quarter of a period (M1): T = 4\times6.0 = 24\,\mathrm{ms}, f = 1/0.024 = 41.7\,\mathrm{Hz} (A1). (b) The antinode amplitude is twice each wave's amplitude: 12/2 = 6.0\,\mathrm{mm} (A1). (c) \lambda = 2\times0.30 = 0.60\,\mathrm{m}; v = f\lambda = 0.60/0.024 = 25\,\mathrm{m\,s^{-1}} (A1). (d) A node is a point of zero amplitude, at rest at every instant (M1). At 6.0\,\mathrm{ms} the two waves cancel everywhere only for that instant; points away from the nodes are crossing equilibrium at their fastest and are displaced again 6.0\,\mathrm{ms} later (A1).

(a) \lambda = 340/500 = 0.680\,\mathrm{m} (M1). Closed at the piston, open at the speaker: L = \lambda/4,\ 3\lambda/4,\ 5\lambda/4, \ldots (M1), giving 0.17, 0.51 and 0.85\,\mathrm{m}; the next, 1.19\,\mathrm{m}, is longer than the tube (A1). (b) 0.85\,\mathrm{m} = 5\lambda/4: the fifth harmonic (A1). Nodes at the piston and 0.34\,\mathrm{m} and 0.68\,\mathrm{m} from it: three (A1). (c) Open at both ends: f_n = nv/2L = n\times170\,\mathrm{Hz} (M1). 500/170 = 2.94 is not a whole number, so there is no resonance at 500\,\mathrm{Hz}; the nearest are 340\,\mathrm{Hz} and 510\,\mathrm{Hz} (A1).

(a) The footsteps are a periodic driving force at about 2\,\mathrm{Hz}, close to the natural frequency (M1); the pushes stay in step with the deck's motion and add energy every cycle, so the amplitude builds up: resonance (A1). (b) Axes labelled amplitude and driving frequency (M1). Before: a tall, narrow peak at about 2.0\,\mathrm{Hz} (A1). After: lower and broader, with its maximum at a slightly lower frequency (A1). (c) The energy removed by damping in each cycle grows with the amplitude; the amplitude stops growing when the energy dissipated per cycle equals the energy the walkers supply per cycle (M1). Dampers dissipate more energy at any given amplitude, so the balance is reached at a smaller amplitude (A1).

(a) Node at the fixed end and antinode at the ring (M1); a quarter of a wavelength, then three quarters (A1). (b) \lambda_1 = 4L = 6.0\,\mathrm{m}, f_1 = 48/6.0 = 8.0\,\mathrm{Hz} (M1); the next allowed mode is the third harmonic, 3\times8.0 = 24\,\mathrm{Hz} (A1). (c) Both ends fixed: f_1 = v/2L = 16\,\mathrm{Hz}, so 16, 32, 48, 64, 80, 96\,\mathrm{Hz} (M1). Before: 8, 24, 40, 56, 72, 88\,\mathrm{Hz}, the odd multiples of 8\,\mathrm{Hz}; after: the even multiples. No frequency is common (A1).

(a) C.1: T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.40/160} = 0.314\,\mathrm{s} (M1); f_0 = 1/T = 3.18\,\mathrm{Hz} (A1). (b) The steady forced oscillation has the driving frequency, 2.0\,\mathrm{Hz} (A1). This is well below 3.18\,\mathrm{Hz}, so the drive is out of step with the natural motion for much of each cycle and little energy is transferred per cycle (A1). (c) f_0 \propto 1/\sqrt{m} (M1): 3.18/\sqrt2 = 2.25\,\mathrm{Hz} (A1). (d) The maximum amplitude is smaller (A1); the peak is broader and sits at a slightly lower frequency (A1).

(a) The microwaves reflect from the metal walls, and incident and reflected waves of the same frequency travel in opposite directions and superpose (M1). The strip dries first at the antinodes, where the field oscillates with the largest amplitude, and remains damp near the nodes (A1). (b) Four gaps: 25.0/4 = 6.25\,\mathrm{cm} between neighbouring antinodes, which is \lambda/2 (M1); \lambda = 12.5\,\mathrm{cm} (A1). C.2: f = c/\lambda = 3.00\times10^{8}/0.125 = 2.40\times10^{9}\,\mathrm{Hz} (A1). (c) B.1: Q = Pt = 700\times90 = 6.3\times10^{4}\,\mathrm{J} (M1); \Delta T = Q/mc = 6.3\times10^{4}/(0.30\times4200) = 50\,\mathrm{K} (A1). (d) The pattern of nodes and antinodes stays fixed in the chamber (M1); moving the belt carries each part through regions of strong and weak field, so the heating evens out (A1).

(a) \lambda = c/f = 3.00\times10^{8}/2.0\times10^{13} = 1.5\times10^{-5}\,\mathrm{m} (A1), in the infrared (A1). (b) \lambda_{\max} = 2.9\times10^{-3}/288 = 1.0\times10^{-5}\,\mathrm{m} (A1). (c) The molecule absorbs strongly only when the radiation's frequency is close to its natural frequency: resonance (M1). Earth's emission is spread around 10^{-5}\,\mathrm{m} and includes 1.5\times10^{-5}\,\mathrm{m}; the Sun's peaks at 2.9\times10^{-3}/5800 = 5.0\times10^{-7}\,\mathrm{m}, far from it (A1). (d) The excited molecules re-emit infrared in all directions, so part returns to the surface (M1); they also pass energy to other molecules in collisions, raising the temperature of the lower atmosphere (A1).

(a) C.1: T = 2\pi\sqrt{L/g} = 2\pi\sqrt{2.45/9.8} (M1) = 3.14\,\mathrm{s} (A1). (b) Pushes every 3.1\,\mathrm{s} arrive at the same point of every swing, in the direction of motion, so each one adds energy: resonance (A1). Pushes every 2.0\,\mathrm{s} drift out of step, sometimes opposing the motion, so there is no steady build-up (A1). (c) A.3: mgh = \tfrac12mv^2 (M1), h = v^2/2g = 2.0^2/19.6 = 0.20\,\mathrm{m} (A1). (d) The amplitude is constant, so the energy lost in each cycle equals the 6.0\,\mathrm{J} supplied in each cycle (M1); average power lost = 6.0/3.14 = 1.9\,\mathrm{W} (A1).