Question:
A wave generates a sinusoidal disturbance and propagates in the positive x-direction with a wavelength of 2 meters. The amplitude of the wave is 0.5 meters, and its frequency is 5 Hz. Determine the wave speed, period, angular frequency, and wave number of this wave.
Answer:
Given: Wavelength (λ) = 2 m Amplitude (A) = 0.5 m Frequency (f) = 5 Hz
To find: Wave speed (v), Period (T), Angular frequency (ω), Wave number (k)
We know that the wave speed is determined by the equation:
v = λ * f
Substituting the given values:
v = 2 m * 5 Hz
v = 10 m/s
Thus, the wave speed of this wave is 10 m/s.
The period (T) of a wave is the reciprocal of frequency (f). So, we can determine it as:
T = 1 / f
T = 1 / 5 Hz
T = 0.2 s
Therefore, the period of this wave is 0.2 seconds.
The angular frequency (ω) is given by the equation:
ω = 2πf
Substituting the given values:
ω = 2π * 5 Hz
ω = 10π rad/s
Hence, the angular frequency of this wave is 10π rad/s.
The wave number (k) can be found using the equation:
k = 2π / λ
Substituting the given value:
k = 2π / 2 m
k = π rad/m
The wave number of this wave is π rad/m.
In summary, the wave characteristics of this wave are as follows: