Celeritas undarum transversarum – problemata et solutiones
1.
If the time interval required to travel from A to B is 2 seconds, determine the speed of the unda transversa.
Notum:
spatium A-B = 6 meters
tempus temporis A-B = 2 seconds
SE busca: Speed of transverse wave
solution:
1 adsum has 1 crest and 1 trough. There is 4 wavelength between A and B. Distance between A and B is 6 meters so that 1 wavelength (λ) = 6 meters / 4 = 1.5 meters.
In time interval required to travel from A to B is 2 seconds so that the time interval required to travel 1 wavelength = period (T) = 2 seconds / 4 = 0.5 seconds.
The equation of the speed of wave :
v = λ f = λ / T
v = speed of wave, λ = adsum, f = frequency, T = ptempus
The speed of wave :
v = 1.5 meters / 0.5 seccondiciones
v = 3 meters/secondicio
2. If the time interval required to travel from A to B is 8 seconds, determine the speed of transverse wave.
Notum:
There are two wavelength.
1 adsum (λ) = 2 x 4 meters = 8 meters
Period (T) = 8 secondiciones / 2 aequalitatem = 4 seconds / adsum
Quaesitum: Speed of wave (V)
solution:
The speed of wave calculated using the equation of the speed of wave :
v = f λ = λ/T
v = 8 meters / 4 seccondiciones
v = 2 meters/secondicio
