Answer:
The frequency of the beat is [tex]f_t = 8Hz[/tex]
Explanation:
From the question we are told that
The speed of the ride [tex]v = 6,2 m/s[/tex]
The frequency of the oboe is [tex]f = 440Hz[/tex]
The temperature outside is [tex]T = 27^oC[/tex]
Generally the speed of sound generated in air is mathematically evaluated as
[tex]v_s = 331 + 0.61 T[/tex]
Substituting value
[tex]v_s = 331 + 0.61 *27[/tex]
[tex]v_s = 347.47 \ m/s[/tex]
The frequency of sound(generated by the musician on he park) getting to the musicians on the unicycle is mathematically evaluated as
[tex]f_a = \frac{v_s }{v_s - v} f[/tex]
substituting values
[tex]f_a = \frac{347.47 }{347.47 - 6.2} * 440[/tex]
[tex]f_a = 448Hz[/tex]
Since the musician on the park is not moving the frequency of sound (from the musicians riding the unicycle )getting to him is = 440Hz
The beat frequency these musician here is mathematically evaluated as
[tex]f_t = f_a - f[/tex]
So [tex]f_t = 448 - 440[/tex]
[tex]f_t = 8Hz[/tex]
