I am trying to figure out the answer for chapter 16 110 a1. I amdoing the proble

Question

I am trying to figure out the answer for chapter 16 110 a1. I amdoing the problem on the books website so the number are slightlydifferent than in the text but the problem should work out thesame. Problem: a wireless transmitting microphone is mounted on asmall platform, which can roll down an incline, away from a speakerthat is mounted at the top of the incline. The speak broadcasts afixed-frequency tone. (a1) The speaker broadcasts a tone that has a frequency of10000Hz and the speed of sound is 343 m/s. At a time of 1.5 sfollowing the release of the platform, the microphone detects afrequency of 9646 Hz. At a time of 3.5 s following the release ofthe platform, the microphone detects a frequency of 9173 Hz. Whatis the acceleration (assumed constant) of the platform? (a1) The speaker broadcasts a tone that has a frequency of10000Hz and the speed of sound is 343 m/s. At a time of 1.5 sfollowing the release of the platform, the microphone detects afrequency of 9646 Hz. At a time of 3.5 s following the release ofthe platform, the microphone detects a frequency of 9173 Hz. Whatis the acceleration (assumed constant) of the platform?

Explanation / Answer

The change in velocity is related to the acceleration of theplatform according to    a =vmike2-vmike1/t-t0 The frequency detected by the microphone at the instant theplatform is released from rest is fo =fs{1-(vmike/v)}   or    vmike =v(1-f0/fs)    for t= 1.5s      vmike1 = (343)(1-9646/10000)              = 12.14 m/s      for t = 3.5s     vmike2 = (343)(1-9173/10000)              = 28.36 m/s the acceleration of the platform is     a = 28.36 m/s - 12.14 m/s / 3.5s-1.5s        = 8.11m/s2 The frequency detected by the microphone at the instant theplatform is released from rest is fo =fs{1-(vmike/v)}   or    vmike =v(1-f0/fs)    for t= 1.5s      vmike1 = (343)(1-9646/10000)              = 12.14 m/s      for t = 3.5s     vmike2 = (343)(1-9173/10000)              = 28.36 m/s the acceleration of the platform is     a = 28.36 m/s - 12.14 m/s / 3.5s-1.5s        = 8.11m/s2              = 28.36 m/s the acceleration of the platform is     a = 28.36 m/s - 12.14 m/s / 3.5s-1.5s        = 8.11m/s2

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