IB Physics C.2: Wave model worked solutions

Vibration parallel to the energy transfer makes the wave longitudinal (A and D fail). The rock particles are the medium: a mechanical wave needs matter to pass the disturbance on, so it cannot cross a vacuum (C fails). Check both halves of every row.

5\times10^{-6}\,\mathrm{m}=5\,\mathrm{\mu m}. Visible light runs from about 4\times10^{-7} to 7\times10^{-7}\,\mathrm{m}, so this wave is about ten times longer than red light: infrared (order 10^{-5}\,\mathrm{m} in the booklet). B treats micrometres as tenths of a micrometre; A goes the wrong way along the spectrum; microwaves are of order 10^{-2}\,\mathrm{m}.

A crest and the next trough are half a cycle apart along the wave: \lambda/2=0.60\,\mathrm{m}. B is 2A, the vertical drop from crest to trough, a distance across the wave, not along it. D is a full wavelength (crest to crest).

One period separates two successive maxima of the same point: T=25.0-5.0=20\,\mathrm{ms}, and f=1/T=1/0.020=50\,\mathrm{Hz}. B takes a maximum-to-minimum gap; C forgets to convert ms to s; D reads the time of the second maximum as the period.

(a) \lambda=v/f=340/425=0.800\,\mathrm{m} (A1). (b) Each oscillation sends out one more wavelength, so the front of the sound is 50\lambda=50(0.800)=40.0\,\mathrm{m} away (A1). Check with time: t=50/425=0.118\,\mathrm{s} and vt=40.0\,\mathrm{m}.

The rope, not the hand, sets the speed: v=f\lambda=2.0(1.5)=3.0\,\mathrm{m\,s^{-1}} before and after. New wavelength \lambda=v/f=3.0/3.0=1.0\,\mathrm{m}. B assumes the speed rises with the frequency, C makes the wavelength grow with the frequency, D multiplies the new wavelength by the old frequency.

Shift the whole pattern a little to the right and read the new displacement at the same place. The crest on P's left moves onto P, so P rises. Q sits at the top of its oscillation, where it stops for an instant before coming down. D moves the rope along with the wave: rope points only move up and down.

\lambda=v/f=20/50=0.40\,\mathrm{m}. The points are 0.10/0.40=\tfrac14 of a wavelength apart, so the phase difference is \tfrac14(2\pi)=\pi/2. C is the half-wavelength answer; D would need the points to be a whole wavelength apart.

I\propto A^2, so I_{\text{new}}=I(1/3)^2=I/9. B treats intensity as proportional to amplitude; C takes a square root instead of squaring.

(a) The power spreads over a sphere: I=P/4\pi r^2=0.50/(4\pi\times4.0^2)=2.49\times10^{-3}\,\mathrm{W\,m^{-2}} (A1). (b) Twice the distance spreads the same power over four times the area: I/4=6.2\times10^{-4}\,\mathrm{W\,m^{-2}} (A1).

The electric field, the magnetic field and the direction of travel are mutually perpendicular. The magnetic field must be horizontal (not up or down, A and D) and at right angles to north (not B): west.

(a) The ribbon is displaced at right angles to the rope and then returns to its original position; it has no net displacement along the rope (A1). (b) As each short section is displaced it pulls on the section ahead and does work on it; energy is handed on section by section to the far end, while every section only oscillates about its own position (A1).

(a) Equilibrium to crest: 2.5\,\mathrm{cm} (A1). (b) Graph 1 gives \lambda=1.20/2=0.60\,\mathrm{m} and graph 2 gives T=0.25\,\mathrm{s} (M1); v=\lambda/T=0.60/0.25=2.4\,\mathrm{m\,s^{-1}} (A1). (c) In 1.0\,\mathrm{s} the point makes 1.0/0.25=4 oscillations and covers 4A=10\,\mathrm{cm} in each (M1): 40\,\mathrm{cm}=0.40\,\mathrm{m}, up and down on the spot, while the wave moves 2.4\,\mathrm{m} along the spring (A1).

(a) First to fifth crest is four gaps: \lambda=6.0/4=1.5\,\mathrm{cm} (A1); 6.0/5 counts crests, not gaps. (b) v=f\lambda=12(0.015)=0.18\,\mathrm{m\,s^{-1}} (A1). (c) The speed depends on the water (its depth), not on the source, so v stays 0.18\,\mathrm{m\,s^{-1}} (M1); \lambda=0.18/16=1.1\times10^{-2}\,\mathrm{m}=1.1\,\mathrm{cm} (A1).

At Q the layers just to the left have positive displacement (pushed right, towards Q) and the layers just to the right have negative displacement (pushed left, towards Q): they crowd together, a compression. At S they move apart (rarefaction). At P and R neighbouring layers are displaced by almost the same amount, so the air there has its normal spacing: the largest displacement is not the largest pressure.

(a) Displacement = actual position minus equilibrium position 5.0n: 0,\ +1.1,\ +1.5,\ +1.1,\ 0,\ -1.1,\ -1.5,\ -1.1,\ 0\,\mathrm{cm}. The largest is 1.5\,\mathrm{cm} (A1). (b) Marker 4: marker 3 is displaced towards it and marker 5 is displaced towards it (M1), so markers 3 and 5 are 7.9\,\mathrm{cm} apart instead of 10\,\mathrm{cm} (A1). (c) Markers 0 and 8 are rarefaction centres (their neighbours move away from them); the next rarefaction is one wavelength on: \lambda=40\,\mathrm{cm} (A1). Not 5.0\,\mathrm{cm}: that is the marker spacing. (d) v=f\lambda=4.0(0.40)=1.6\,\mathrm{m\,s^{-1}} (A1).

Intensity ratio =2\times(2.0/4.0)^2=0.50; amplitude ratio =\sqrt{0.50}=0.71. B stops at the intensity ratio; A ignores the doubled power; D applies the inverse square law to the amplitude and the power ratio to it as well.

(a) The same disturbance travels by two media at different speeds, so it arrives at two different times (A1). (b) t_{\text{air}}=680/340=2.00\,\mathrm{s} and t_{\text{steel}}=680/5100=0.133\,\mathrm{s} (M1); interval =1.87\,\mathrm{s} (A1). Subtracting the speeds first gives nothing useful. (c) The hammer sets the frequency in both: \lambda_{\text{steel}}=5100/850=6.0\,\mathrm{m} (A1), \lambda_{\text{air}}=340/850=0.40\,\mathrm{m} (A1).

(a) Any two of: sound is longitudinal while microwaves are transverse; sound needs a material medium while microwaves also travel through a vacuum; in sound, particles of the air oscillate, in a microwave the electric and magnetic fields oscillate; sound travels at about 340\,\mathrm{m\,s^{-1}} in air, microwaves at about 3.0\times10^{8}\,\mathrm{m\,s^{-1}} (A1 each). (b) \lambda=c/f=3.00\times10^{8}/2.45\times10^{9}=0.122\,\mathrm{m} (A1). (c) The magnetic field is perpendicular to both the vertical electric field and the eastward travel: horizontal, along the north–south line (A1).

0.60/0.80=\tfrac34 of a cycle. Starting at zero and moving down, M reaches the trough after \tfrac14 cycle, zero (moving up) after \tfrac12 and the crest after \tfrac34. It never moves along the rope, so C is wrong; B stops after a quarter cycle; D after half a cycle.

(a) I=P/4\pi r^2=5.0\times10^{3}/(4\pi(1.0\times10^{4})^2)=3.98\times10^{-6}\,\mathrm{W\,m^{-2}} (A1). (b) I\propto1/r^2: one quarter at twice the distance, 20\,\mathrm{km} (A1). (c) I\propto A^2 (M1), so the amplitude falls by \sqrt{1/4}=0.50 (A1); a quarter of the amplitude would need I/16.

(a) 2.0\,\mathrm{\mu m}=2.0\times10^{-6}\,\mathrm{m} (A1). (b) f=1/T=1/(0.20\times10^{-3})=5.0\times10^{3}\,\mathrm{Hz} (A1). (c) The disturbance needs 0.48\,\mathrm{ms} to cover 2.4\,\mathrm{m}: v=2.4/0.48\times10^{-3}=5.0\times10^{3}\,\mathrm{m\,s^{-1}} (A1). (d) \lambda=v/f=1.0\,\mathrm{m} (A1).

(a) The same crest moved 0.35-0.20=0.15\,\mathrm{m} (M1) in 0.050\,\mathrm{s}: v=3.0\,\mathrm{m\,s^{-1}} (A1). Assuming a quarter-wavelength shift (0.20\,\mathrm{m}) gives 4.0\,\mathrm{m\,s^{-1}} and loses both marks. (b) f=v/\lambda=3.0/0.80=3.75\,\mathrm{Hz} (A1). (c) One more wavelength would look identical: shift 0.95\,\mathrm{m}, v=0.95/0.050=19\,\mathrm{m\,s^{-1}} (A1). (d) At t=0 the point is at zero displacement with the crest at 0.20\,\mathrm{m} to its left; the crest is coming towards it, so it is moving upwards (A1). (e) The crest must travel 0.20\,\mathrm{m} (M1): t=0.20/3.0=0.067\,\mathrm{s} (A1), a quarter of the period 0.267\,\mathrm{s}, as it should be.

(a) \lambda=vT=2.0(0.40)=0.80\,\mathrm{m} (A1). (b) The front has travelled vt=1.2\,\mathrm{m}, so the rope beyond 1.2\,\mathrm{m} is still flat (A1). Behind the front, a point at distance x repeats what the end did x/v earlier; the end has done 1.5 cycles, so the rope shows 1.5\lambda: y=+4.0\,\mathrm{cm} at x=0.20 and 1.0\,\mathrm{m}, -4.0\,\mathrm{cm} at 0.60\,\mathrm{m}, zero at 0, 0.40, 0.80 and 1.2\,\mathrm{m} (A1 shape and wavelength, A1 correct phase). The part just behind the front is displaced upwards, because the end's first movement was upwards. (c) The crest at 0.20\,\mathrm{m} is travelling towards x=0.40\,\mathrm{m}, which is at zero, so this point is moving upwards (A1). (d) In one period a point moves 4A=16\,\mathrm{cm} (M1): average speed 0.16/0.40=0.40\,\mathrm{m\,s^{-1}}, a fifth of the wave speed; the two speeds measure different motions (A1).

(a) Same amplitude means the same intensity (M1): 4P/4\pi r^2=P/4\pi(12)^2, so r^2=4(144) and r=24\,\mathrm{m} (A1). (b) Amplitude 1.5a needs intensity 1.5^2=2.25 times A's (M1): 4(12/r)^2=2.25, r=12\times2/1.5=16\,\mathrm{m} (A1). (c) I=0.80/(4\pi\times16^2)=2.49\times10^{-4}\,\mathrm{W\,m^{-2}} (A1). (d) Some energy is absorbed by the air, the ground and the site, or reflected away, so less than the full power crosses the sphere; the siren may also radiate more strongly in some directions (A1).

(a) 600/40=15\,\mathrm{m\,s^{-1}} (A1). (b) A person stands while the pattern moves 15(0.80)=12\,\mathrm{m} (M1), so about 12/0.50=24 people stand at once (A1). (c) People move vertically, the pattern horizontally: transverse (A1). (d) In a rope each section pulls on the next and does work on it, so energy travels from the source along the rope (M1). In the stadium each person stands up using their own energy because they see the pattern coming; no person pushes the next, so the pattern moves but no energy is carried along the stands (A1).

(a) \lambda=340/400=0.85\,\mathrm{m} (A1). (b) Delay =0.34/340=1.0\times10^{-3}\,\mathrm{s} (A1). (c) The sensors are 0.34/0.85=0.40 of a wavelength apart (M1): \phi=0.40(2\pi)=0.8\pi=2.51\,\mathrm{rad} (A1). Equivalently delay/period =1.0/2.5=0.40. (d) The speed is fixed by the air, so raising f shortens \lambda (M1). The traces first agree when the separation is one whole wavelength: \lambda=0.34\,\mathrm{m}, f=340/0.34=1.0\times10^{3}\,\mathrm{Hz} (A1). (e) Points one wavelength apart are in phase: another compression centre (A1).

(a) A.1: the echo covers the depth twice, 2d=vt (M1), d=1500(0.12)/2=90\,\mathrm{m} (A1). (b) \lambda=1500/3.0\times10^{4}=0.050\,\mathrm{m} (A1). (c) N=ft=3.0\times10^{4}(2.0\times10^{-3})=60 wavelengths (A1); length =60(0.050)=3.0\,\mathrm{m}, the same as vt (A1). (d) C.3: the source fixes the frequency, so it stays 30\,\mathrm{kHz} when the wave changes medium (A1); \lambda=340/3.0\times10^{4}=1.1\times10^{-2}\,\mathrm{m} (A1).

(a) f=v/\lambda=1.2/2.4=0.50\,\mathrm{Hz} (A1), T=1/f=2.0\,\mathrm{s} (A1). (b) C.1: \omega=2\pi f=\pi\,\mathrm{rad\,s^{-1}} (M1); a_{\max}=\omega^2x_0=\pi^2(0.060)=0.59\,\mathrm{m\,s^{-2}} (A1). (c) G is 0.60/2.4=\tfrac14 wavelength further on, so it is a quarter cycle behind F (M1). A quarter cycle before its own highest point, G is at zero displacement and moving upwards (A1). (d) I\propto A^2: (4.0/6.0)^2=0.44 (A1).

(a) B.1: \lambda_{\max}=2.9\times10^{-3}/5800=5.0\times10^{-7}\,\mathrm{m} (A1), visible (A1). (b) f=c/\lambda=3.00\times10^{8}/5.0\times10^{-7}=6.0\times10^{14}\,\mathrm{Hz} (A1). (c) The power spreads over a sphere of radius 1.5\times10^{11}\,\mathrm{m} (M1): I=L/4\pi d^2=3.8\times10^{26}/(4\pi(1.5\times10^{11})^2)=1.34\times10^{3}\,\mathrm{W\,m^{-2}} (A1). (d) I\propto1/d^2: 1.34\times10^{3}/1.52^2=5.8\times10^{2}\,\mathrm{W\,m^{-2}} (A1). (e) Light is an electromagnetic wave: oscillating electric and magnetic fields sustain each other, so no material medium is needed (A1).

(a) There and back is 2L=1.30\,\mathrm{m} (M1): t=1.30/286=4.5\times10^{-3}\,\mathrm{s} (A1). (b) C.4: nodes at both ends with one loop between them, so \lambda=2L=1.30\,\mathrm{m} (A1); f=v/\lambda=286/1.30=220\,\mathrm{Hz} (A1). (c) The frequency is the same in the air: \lambda=340/220=1.55\,\mathrm{m} (A1). (d) The string's points are displaced at right angles to the string, along which energy travels (A1). The vibrating string pushes and pulls the air beside it, so air layers move back and forth along the direction in which the sound travels away from the string (A1).