IGCSE Physics 0625: Wavelength, Frequency and Speed of a Conveyor Belt Ripple Wave

A bottling-factory technician notices that the flat rubber conveyor belt carrying glass bottles has developed a travelling ripple fault along its surface, caused by a worn support roller. The ripple pattern moves continuously along the length of the belt.

(a) Using a strobe light, the technician freezes the belt's motion and measures the distance from one ripple crest to the next crest as $0.24\text{ m}$ , and the height of each ripple crest above the flat resting level of the belt as $3.0\text{ mm}$ . State the wavelength and the amplitude of this ripple wave. [2]

(b) Using a stopwatch, the technician then times $15$ complete ripple cycles passing a fixed point on the belt in $6.0\text{ s}$ . Calculate the frequency of the ripple wave. [2]

(d) State whether this wave on the belt is transverse or longitudinal, and justify your answer in terms of the direction the belt material moves compared with the direction the wave travels. [2]

Part (a): Reading off wavelength and amplitude

The wavelength of a wave is the distance between one point on the wave and the identical point on the next full cycle. Here, the distance from one crest to the next crest:

$\lambda = 0.24\text{ m}$

The amplitude is the maximum height of the wave above its rest (undisturbed) position. Here, the height of a ripple crest above the flat resting level of the belt:

$\text{amplitude} = 3.0\text{ mm}$

Part (b): Calculating the frequency

Frequency is the number of complete waves passing a fixed point per second. The technician counted $15$ complete cycles in $6.0\text{ s}$ :

$f = \frac{\text{number of cycles}}{\text{time}} = \frac{15}{6.0\text{ s}}$

$f = 2.5\text{ Hz}$

Part (c): Calculating the wave speed

The wave equation links speed, frequency and wavelength:

$v = f\lambda$

Substituting the frequency from part (b) and the wavelength from part (a):

$v = 2.5\text{ Hz} \times 0.24\text{ m}$

$v = 0.60\text{ m/s}$

Part (d): Classifying the wave

To classify a wave, compare the direction the material itself vibrates with the direction the wave travels:

The belt’s rubber surface moves up and down, perpendicular to the belt.
The ripple wave itself travels along the length of the belt.

Since the vibration is perpendicular to the direction of wave travel, this is a transverse wave.

Final answers

(a) Wavelength $= \boxed{0.24\text{ m}}$ ; amplitude $= \boxed{3.0\text{ mm}}$
(b) Frequency $= \boxed{2.5\text{ Hz}}$
(c) Wave speed $= \boxed{0.60\text{ m/s}}$
(d) Transverse, the belt vibrates perpendicular to the direction the wave travels