Waves for Cambridge IGCSE Physics

Reading cue. The 2.5 cm dimension runs crest to next crest, not crest to trough, and the scale line is the undisturbed water level, so amplitude is measured from that line rather than trough to crest.

1. aDetermine Determine the total horizontal distance occupied by the complete waves drawn in Fig. 2.1.

   recall2 marks

   Check answer · 2 marks

   1. four complete wavelengths are drawn
   2. 4 × 2.5 = 10 cm

2. bDetermine The pattern drawn in Fig. 2.1 takes 0.50 s to travel a distance equal to the whole marked scale. Determine the frequency of the waves.

   routine2 marks

   Check answer · 2 marks

   1. speed = 10 cm / 0.50 s = 20 cm/s
   2. f = v/λ = 20 / 2.5 = 8.0 Hz

3. cDescribe A small cork floats on the water at the 5.0 cm mark on the scale in Fig. 2.1. Describe how the cork moves as the waves pass, and state how far it travels along the scale.

   demanding3 marks

   Check answer · 3 marks

   1. the cork moves up and down about the undisturbed water level, at right angles to the arrow showing the direction of travel
   2. it completes one full oscillation each time a whole wave passes it
   3. it travels no distance along the scale, because the wave transfers energy along the tank without carrying the water along with it

4. dExplain A student says that doubling the frequency of the dipper will double the number of complete waves fitting into the 10 cm scale in Fig. 2.1, but will leave the speed of a crest unchanged. Explain whether the student is correct.

   top of the paper4 marks

   Check answer · 4 marks

   1. the speed of the water waves is fixed by the depth of the water, not by the dipper, so the crest speed is unchanged
   2. since v = fλ and v is fixed, doubling f halves the wavelength from 2.5 cm to 1.25 cm
   3. the 10 cm scale then holds 8 complete waves instead of 4
   4. the student is correct on both counts

Reading cue. The 165 m dimension line is the one-way distance to the wall, so the sound covers it twice; the figure deliberately draws no sound path for you to count.

1. aDetermine Determine the total distance travelled by the sound between the clap and the student hearing the echo.

   recall2 marks

   Check answer · 2 marks

   1. the sound travels from the student to the wall and back again
   2. 2 × 165 = 330 m

2. bDetermine The student walks towards the wall until the echo returns 0.40 s after she claps. Taking the speed of sound in air as 330 m/s, determine her new distance from the wall.

   routine2 marks

   Check answer · 2 marks

   1. total path = 330 × 0.40 = 132 m
   2. distance to the wall = 132/2 = 66 m

3. cDetermine Back at the position drawn in Fig. 5.1 she times the echo at 1.0 s after the clap, but her timing could be wrong by up to 0.20 s either way. Determine the largest and the smallest speeds of sound consistent with the 165 m marked in the figure.

   demanding3 marks

   Check answer · 3 marks

   1. path length = 2 × 165 = 330 m
   2. largest speed = 330/0.80 = 412.5 m/s, that is 410 m/s to 2 significant figures
   3. smallest speed = 330/1.20 = 275 m/s

4. dSuggest Suggest whether she would obtain a better value for the speed of sound by doubling the 165 m marked in Fig. 5.1 or by buying a stopwatch that reads to 0.01 s, and give reasons.

   top of the paper3 marks

   Check answer · 3 marks

   1. the timing error is set by her reaction time, not by how finely the stopwatch reads, so a more precise stopwatch changes almost nothing
   2. doubling the distance doubles the echo time while the reaction error stays the same size
   3. the error is then a smaller fraction of the measured time, so moving further from the wall is the better improvement

Reading cue. The 30° arc is between a wavefront and the boundary, not between a ray and a normal, and the shallow half of the tank is left blank on purpose: you must say what belongs there.

1. aState The shallow region of Fig. 7.1 is drawn empty. State how the spacing of the wavefronts there compares with the spacing of those drawn in the deep region, and state what happens to the frequency of the waves as they cross the boundary.

   recall2 marks

   Check answer · 2 marks

   1. the wavefronts in the shallow region are closer together, because the waves travel more slowly there
   2. the frequency is unchanged

2. bDetermine The dipper makes twelve complete waves in 1.5 s, producing the twelve wavefronts drawn in Fig. 7.1. Determine the frequency of the waves, and hence their wavelength in the shallow region, where they travel at 9.6 cm/s.

   routine2 marks

   Check answer · 2 marks

   1. f = 12/1.5 = 8.0 Hz
   2. λ = v/f = 9.6/8.0 = 1.2 cm

3. cDetermine The twelve wavefronts drawn in the deep region of Fig. 7.1 are evenly spaced, one wavelength of 1.8 cm apart. Determine the distance from the first of them to the twelfth, and determine the distance those same twelve wavefronts occupy once all of them have crossed into the shallow water.

   demanding3 marks

   Check answer · 3 marks

   1. there are 11 wavelengths between the first wavefront and the twelfth, not 12
   2. distance in the deep region = 11 × 1.8 = 19.8 cm
   3. the spacing in the shallow region is 1.2 cm, so the distance = 11 × 1.2 = 13.2 cm

4. dExplain The dipper is now vibrated twice as fast and nothing else is altered. Explain what happens to the wavelength in each region of Fig. 7.1, and explain why the wavefronts in the shallow water still make the same angle with the boundary as before.

   top of the paper4 marks

   Check answer · 4 marks

   1. the speed in each region is fixed by the depth of the water there, so neither speed changes
   2. λ = v/f, so doubling the frequency halves each wavelength: 0.90 cm in the deep region and 0.60 cm in the shallow region
   3. the ratio of the two wavelengths is 0.60/0.90, the same value as 1.2/1.8 before
   4. the change of direction at the boundary is set by that ratio and not by the frequency, so the angle the wavefronts make with the boundary is unchanged