Given:
Waves A, B, and C travel 12.0 meters in 2.0 seconds through the given medium.
To find:
The speed, frequency, period, wavelength, and amplitude of wave A.
Explanation:
As the wave covers a distance of 12 m in 2 seconds, the speed "v" of the wave can be calculated as:
[tex]v=\frac{d}{t}[/tex]
Here, d is the distance covered by the wave and t is the time required to cover the distance.
Substituting the d = 12 m and t = 2 s in the above equation, we get:
[tex]v=\frac{12\text{ m}}{2\text{ s}}=6\text{ m/s}[/tex]
The speed of the wave A is 6 m/s.
The frequency of a wave is defined as the number of wave cycles occurring per second.
From the above graph, we observe that there are a total of four wave cycles occurring in 2 s. Thus, the frequency of wave A can be calculated as:
[tex]f=\frac{N}{t}[/tex]
Here, f is the frequency of the wave, N is the number of times the wave occurs and t is the time.
Substituting N = 4 and t = 2 s in the above equation, we get:
[tex]f=\frac{4}{2\text{ s}}=2\text{ Hz}[/tex]
The frequency of wave A is 2 Hz.
The period "T" of the wave is related to the frequency "f" as:
[tex]T=\frac{1}{f}[/tex]
Substituting the values in the above equation, we get:
[tex]T=\frac{1}{2\text{ Hz}}=0.5\text{ s}[/tex]
The period of wave A is 0.5 s.
The wavelength λ of the wave is related to its speed "v" and frequency "f" as:
[tex]\begin{gathered} v=f\lambda \\ \\ \lambda=\frac{v}{f} \end{gathered}[/tex]
Substituting the values in the above equation, we get:
[tex]\lambda=\frac{6\text{ m/s}}{2\text{ Hz}}=3\text{ m}[/tex]
The wavelength of wave A is 3 m.
The amplitude of the wave is the maximum distance or displacement of the point on a vibrating body measured from the equilibrium position. Thus, from the given graph, we observe that the amplitude of wave A is 2 m.
Final answer:
The speed of wave A is 6 m/s.
The frequency of wave A is 2 Hz.
The time period of wave A is 0.5 s.
The wavelength of wave A is 3 m.
The amplitude of wave A is 2 m.
