   Chapter 14, Problem 33P

Chapter
Section
Textbook Problem

A tuning fork vibrating at 512 Hz falls from rest and accelerates at 9.80 m/s2. How far below the point of release is the tuning fork when waves of frequency 485 Hz reach the release point?

To determine
The total distance fallen before the 485 Hz sound reaches the observer

Explanation

Given Info: Frequency of vibrating tuning fork is 512 Hz, frequency of tuning fork releases at the point is 485Hz and acceleration is 9.8m/s2 . Observer is at rest vo=0 .

Formula to calculate the observed frequency is,

fo=fs(v+vovvs)

• vs is the speed of falling turning fork
• vo is the speed of the observer,
• fs is the frequency of vibrating turning fork
• fo is the frequency of tuning fork observed at the release point,

Rearrange the expression in terms of vs

fo=fs(v+vovvs)(vvs)fo=fs(v+vo)(vvs)=fs(v+vo)fovs=vfs(v+vo)fo

The vibrating turning fork recedes from the stationary observer. The observer is at rest, vo=0 .

The source is moving away from the observer, vs=vs .

Substitute 512 Hz for fs 485 Hz for fo , 0 for vo , 343m/s for v, and vs for vs in the above expression to calculate vs .

vs=(343m/s512Hz(343m/s+0)485Hz)=19.09m/svs19.1m/s

Formula to calculate the distance it has fallen from rest before reaching this speed,

Δy1=vs2u22ay

• u is the initial speed of the tuning fork,
• ay is the acceleration of the tuning fork,

Substitute 19

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