Answer:
The two points are 0.33 m far apart
The phase difference between two displacements occurring at times separated by 0.01 s = 72°
Explanation:
Given that:
frequency of the wave (n) = 20 Hz
the velocity of the wave = 80 m/s
We can determine the wavelength ([tex]\lambda[/tex]) = [tex]\frac{velocity of the wave (v)}{frequency of the wave (n)}[/tex]
([tex]\lambda[/tex]) = [tex]\frac{80}{20}[/tex]
([tex]\lambda[/tex]) = 4 m
In order to determine how far apart the two points whose displacements are 30°apart in phase; we have the phase difference (∅) at 30° = [tex]\frac{\pi }{6}[/tex] rads
For path difference ([tex]\lambda[/tex]) , there is phase difference of 2π
Therefore, we can say that:
∅ = [tex]\frac{2\pi x}{\lambda}[/tex]
x = [tex]\frac{\theta \lambda}{2 \pi}[/tex]
x = [tex]\frac{\pi}{6} *\frac{4}{2 \pi}[/tex]
x = 0.33 m
b)
At a given point, what is the phase difference between two displacements occurring at times separated by 0.01 s
If (t) = 0.01 seconds
Then; path difference (x) = v × t = 80 × 0.01
= 0.8 m
phase difference can therefore be calculated as:
= [tex]\frac{0.8*360}{4}[/tex]
= 72°
