Answer:
The value is [tex]v_s = 1.394 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]
The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s [/tex]
Generally the frequency of the note played by the first player is mathematically represented as
[tex]f_1 = f_2 + f_b[/tex]
=> [tex]f_1 = 490 + 2.0 [/tex]
=> [tex]f_1 = 492 Hz[/tex]
From the relation of Doppler Shift we have that
[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]
Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]
So
[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]
=> [tex]v_s = 1.394 \ m/s[/tex]
