Question

section b: 55 marks

1. a) define the term progressive waves.

(1 mark)

b) figure shows air molecules in front of a hollow, wooden box b set vibrating by a tuning fork.

i) what is the name given to this kind of wave?
(1 mark)

ii) name sections x and y?
(2 marks)

iii) a wave front takes 0.03 seconds to travel from x to z. the velocity of sound in air is 330m/s. calculate the frequency of wave produced.
(3 marks)

c) a mine - worker stands between two vertical cliffs 400m from the nearest cliff. the cliffs are y distances apart. every time he strikes the rock once, he hears two echoes; the first one after 2.5s, while the second follows 2s later. from this information, calculate;
i) the speed of the sound in air.
(3 marks)

ii) the value of y.
(3 marks)

page 5 of 10

Explanation:

Part a)

Define progressive waves: Waves transferring energy without net matter movement.

Part b)

i)

Identify wave type: Longitudinal (air molecules vibrate parallel to wave direction).

ii)

Name sections: X = Compression (molecules close), Y = Rarefaction (molecules spread).

iii)

Step1: Calculate distance X-Z
Distance = velocity × time = $330 \, \text{m/s} × 0.03 \, \text{s} = 9.9 \, \text{m}$

Step2: Count wavelengths (X to Z: 3 full waves)
Wavelength $\lambda = \frac{9.9 \, \text{m}}{3} = 3.3 \, \text{m}$

Step3: Calculate frequency
Frequency $f = \frac{v}{\lambda} = \frac{330}{3.3} = 100 \, \text{Hz}$

Part c)

I)

Step1: First echo (nearest cliff: 400m round trip = 800m)
Speed $v = \frac{\text{distance}}{\text{time}} = \frac{800 \, \text{m}}{2.5 \, \text{s}} = 320 \, \text{m/s}$

II)

Step1: Time for second echo = $2.5 + 2 = 4.5 \, \text{s}$
Step2: Distance for second echo = $v × t = 320 × 4.5 = 1440 \, \text{m}$ (round trip)
Step3: Distance to far cliff = $\frac{1440}{2} = 720 \, \text{m}$
Step4: Total distance Y = 400 + 720 = 1120 m

Answer:

a) A wave that transfers energy through a medium without net movement of the medium itself.
b) i) Longitudinal wave
ii) X: Compression; Y: Rarefaction
iii) 100 Hz
c) I) 320 m/s
II) 1120 m