Example 1
The diagram below shows a particular stationary wave pattern produced by sound in a closed tube.
The length L of the tube is 30cm.
aDraw an arrow to show the oscillations of the air particles at the open end of the tube.
bDetermine thewavelength of sound.
cThe speed of sound is 340ms−1. Calculate the frequency of sound.
aSound is a longitudinal wave. The double-headed arrow shows the oscillations of the air particles.
bThe separation between adjacent nodes is equal to .
Therefore:cm
cm /
cThe speed of sound is given by the wave equation: v = f.
Therefore:
340 = f×0.40
Hz
COAS Physics 1 Teacher ResourcesOriginal material © Cambridge University Press 2005, 20081
18 Worked examples
Example 2
A microwave transmitter is placed in front of a metal plate. The microwaves are reflected off the plate. A stationary (standing) wave is formed in the space between the plate and the transmitter. A detector is moved between the transmitter and the plate. Four nodes are detected in a distance of 8.0cm. Calculate the wavelength and frequency of the microwaves.
There are three ‘loops’ of the stationary wave.
Therefore:
m ≈ 5.3cm
The wavelength of the microwaves is 5.3cm.
The frequency f of the microwaves can be found using c = f.
Therefore:3.0×108 = f×5.33×10−2
Hz (5.6GHz)
The frequency of the microwaves is 5.6GHz. /
Example 3
The diagram below shows a dry plastic tube containing some fine powder (dust). One end of the tube is closed. At the open end, a loudspeaker vibrating at a frequency of 4.5 kHz is positioned, as shown in the diagram.
At this frequency, the powder within the tube forms heaps at the nodes. Calculate the speed of sound.
The separation between adjacent nodes is equal to half of a wavelength.Therefore:
cmcm /
The wavelength of the sound within the tube is 0.075m.
The speed of sound is given by the wave equation: v = f.
Therefore:
v = 4500×0.075
v≈ 340ms−1
The speed of sound is 340ms−1.
COAS Physics 1 Teacher ResourcesOriginal material © Cambridge University Press 2005, 20081