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A wireless transmitting microphone is mounted on a small platform that can roll down an incline, directly away from a loudspeaker that is mounted at the top of the incline. The loudspeaker broadcasts a tone that has a frequency of 1.000 × 104 Hz, and the speed of sound is 343 m/s. At a time of 1.5 s following the release of the platform, the microphone detects a frequency of 9952 Hz. At a time of 3.5 s following the release of the platform, the microphone detects a frequency of 9888 Hz. What is the acceleration a (assumed constant) of the platform

Respuesta :

Answer:

Acceleration, a = 1.1 m/s²

Explanation:

Frequency of source, [tex]f_s=10^4\ Hz[/tex]

Speed of sound, v = 343 m/s

Frequency of observer, [tex]f_o=9952\ Hz[/tex]

We know that acceleration a of the platform is :

[tex]a=\dfrac{v_2-v_1}{t_2-t_1}[/tex]..............(1)

The frequency detected by the microphone is :

At 1.5 seconds,

[tex]f_o=f_s(1-\dfrac{v_2}{v})[/tex]

[tex]v_2=v(1-\dfrac{f_o}{f_s})[/tex]

[tex]v_2=343\times (1-\dfrac{9952}{10000})[/tex]

[tex]v_2=1.64\ m/s[/tex]

At 3.5 seconds,

[tex]f_o=f_s(1-\dfrac{v_2}{v})[/tex]

[tex]v_2=v(1-\dfrac{f_o}{f_s})[/tex]

[tex]v_2=343\times (1-\dfrac{9888}{10000})[/tex]

[tex]v_2=3.84\ m/s[/tex]

So, equation (1) becomes :

[tex]a=\dfrac{3.84-1.64}{3.5-1.5}[/tex]

[tex]a=1.1\ m/s^2[/tex]

So, the acceleration of the platform is 1.1 m/s². Hence, this is the required solution.