Answers · Waves in a ripple tank

Waves in a ripple tank — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. VK1 · Final check

Model answer. (a) A wave transfers energy. It does not transfer matter: the material it travels through vibrates about its own place and ends up where it started. (b) In a transverse wave the direction of vibration is at right angles to the direction of propagation, as in light, water waves and seismic S-waves. In a longitudinal wave it is parallel to the direction of propagation, as in sound and seismic P-waves. (c) The wavelength is the distance from one crest to the next. The amplitude is the greatest distance a point moves from its rest position, from the rest line to a crest. (d) The frequency is the number of waves each second: f = 20 ÷ 5.0 s = 4.0 Hz. Then λ = v ÷ f = 0.24 m/s ÷ 4.0 Hz = 0.060 m. (e) Switch on the dipper and take a still picture of the bright lines. Lay a ruler along the direction of travel and measure across many wavelengths, from one bright line to another several lines further on. Divide by the number of gaps between those lines, which is one less than the number of lines.

A full answer includes

(a) I wrote that a wave transfers energy.
(a) I wrote that it does not transfer matter.
(b) I wrote that in a transverse wave the vibration is at right angles to the direction of propagation.
(b) I wrote that in a longitudinal wave the vibration is parallel to the direction of propagation.
(b) I gave a transverse example: light (any electromagnetic wave), water waves or seismic S-waves.
(b) I gave a longitudinal example: sound or seismic P-waves.
(c) I wrote that the wavelength is the distance from one crest to the next.
(c) I wrote that the amplitude is the distance from the rest position to a crest.
(c) I did not define the amplitude as the distance from a trough to a crest.
(d) I wrote f = 20 ÷ 5.0 s = 4.0 Hz.
(d) I wrote λ = v ÷ f = 0.24 m/s ÷ 4.0 Hz.
(d) I wrote 0.060 m.
(e) I wrote that I take a still picture of the bright lines.
(e) I wrote that I measure across many wavelengths with a ruler.
(e) I wrote that I divide by the number of gaps between the lines.
(e) I did not divide by the number of lines.

Also accept

(a) energy, not matter (b) right angles / parallel, with an example of each (c) crest to crest; rest position to crest (d) 0.060 m (e) still picture; measure many wavelengths; divide by the number of gaps

Watch for

“The amplitude is the height from a trough to a crest.” — That height is twice the amplitude. The amplitude is from the rest position to a crest.
“The frequency is 5.0 s ÷ 20 = 0.25 Hz.” — That is the time for one wave. The frequency is the number of waves each second.

2. VK2 · Final check

Model answer. (a) 35°. The reflected wavefronts make the same angle with the barrier as the arriving ones. (b) Water waves travel faster in deeper water, so the speed increases. The frequency stays the same, so the wavelength increases too (λ = v ÷ f). (c) Beyond the gap the wavefronts become curved and spread out into the water on both sides. The wavelength stays the same, and so do the frequency and the speed, because the water beyond the gap is the same. (d) Waves spread out in semicircles when the gap is about the same size as the wavelength, so the wavelength is about 3.5 cm. (e) Ship waves: λ = v ÷ f = 3.0 m/s ÷ 0.50 Hz = 6.0 m. Ripples: λ = 0.56 m/s ÷ 2.8 Hz = 0.20 m. The ship waves bend further: their wavelength is much longer, and the longer the wavelength, the further waves bend round an edge.

A full answer includes

(a) I wrote 35°.
(a) I did not write 55°.
(b) I wrote that the speed increases.
(b) I wrote that the wavelength increases.
(b) I did not write that the frequency changes.
(c) I wrote that the wavefronts become curved and spread out beyond the gap.
(c) I gave one quantity that stays the same: the wavelength, the frequency or the speed.
(d) Supplement: I wrote that semicircles beyond a gap mean the gap is about one wavelength wide.
(d) Supplement: I wrote about 3.5 cm.
(d) Supplement: I did not give a wavelength much smaller than the gap.
(e) Supplement: I used λ = v ÷ f for each set of waves.
(e) Supplement: I wrote 6.0 m for the ship waves.
(e) Supplement: I wrote 0.20 m for the ripples.
(e) Supplement: I wrote that the ship waves bend further.
(e) Supplement: I wrote that longer waves bend further round an edge.
(e) Supplement: I did not choose the ripples because their frequency is higher.

Also accept

(a) 35° (b) both increase (c) curved, spreading out; wavelength (or frequency or speed) unchanged (d) about 3.5 cm (e) 6.0 m and 0.20 m: the ship waves, with the longer wavelength

Watch for

“The reflected wavefronts make 55° with the barrier.” — 55° is the angle between the barrier and the direction the waves travel. The reflected wavefronts make the same angle with the barrier as the arriving ones.
“Refraction changes the frequency of the waves.” — The frequency stays the same; refraction changes the speed, the wavelength and the direction.
“The ripples bend further, because their frequency is higher.” — How far waves bend round an edge depends on their wavelength. The ripples are only 0.20 m long, far shorter than the 6.0 m ship waves, so they bend less.