Question

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

Answer 1

Acceleration, a = 1.1 m/s²

Step-by-step explanation:

Frequency of source,
`f_s=10^4\ Hz`

Speed of sound, v = 343 m/s

Frequency of observer,
`f_o=9952\ Hz`

We know that acceleration a of the platform is :

`a=(v_2-v_1)/(t_2-t_1)`..............(1)

The frequency detected by the microphone is :

At 1.5 seconds,

`f_o=f_s(1-(v_2)/(v))`

`v_2=v(1-(f_o)/(f_s))`

`v_2=343* (1-(9952)/(10000))`

`v_2=1.64\ m/s`

At 3.5 seconds,

`f_o=f_s(1-(v_2)/(v))`

`v_2=v(1-(f_o)/(f_s))`

`v_2=343* (1-(9888)/(10000))`

`v_2=3.84\ m/s`

So, equation (1) becomes :

`a=(3.84-1.64)/(3.5-1.5)`

`a=1.1\ m/s^2`

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