Answer:
Explanation:
From the question we are told that
The length of the wire is [tex]L = 75.0cm = \frac{75}{100} = 0.75 \ m[/tex]
The mass of the wire is [tex]m = 2.25 \ g = \frac{2.25}{1000} = 0.00225 \ kg[/tex]
The tension is [tex]T = 400 \ N[/tex]
The frequency of the beat heard by the second student is
[tex]f_b = 8.30\ beat/second[/tex]
The speed of the wave generated by the vibration of the wire is mathematically represented as
[tex]v = \sqrt{\frac{TL}{m}}[/tex]
substituting values
[tex]v = \sqrt{\frac{400 *0.75}{0.00225}}[/tex]
[tex]v = 365.15 m/s[/tex]
The wire is vibrating in its third harmonics so the wavelength is
[tex]\lambda = \frac{2L}{3}[/tex]
substituting values
[tex]\lambda = \frac{2*0.75}{3}[/tex]
[tex]\lambda = 0.5 \ m[/tex]
The frequency of this vibration is mathematically represented as
[tex]f = \frac{v}{\lambda }[/tex]
substituting values
[tex]f = \frac{365.15}{0.5 }[/tex]
[tex]f = 730.3 Hz[/tex]
The speed of the second student (Observer) is mathematically represented as
[tex]v_o = [\frac{f_b}{2f} ] * v[/tex]
substituting values
[tex]v_o = [\frac{8.30}{2* 730.3} ] * 365.15[/tex]
[tex]v_o = 2.08 \ m/s[/tex]
