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Unformatted text preview: 97. The siren is between you and the cliﬀ, moving away from you and towards the cliﬀ. Both “detectors” (you
and the cliﬀ) are stationary, so vD = 0 in Eq. 18-47 (and see the discussion in the textbook immediately
after that equation regarding the selection of ± signs). The source is the siren with vS = 10 m/s. The
problem asks us to use v = 330 m/s for the speed of sound.
(a) With f = 1000 Hz, the frequency fy you hear becomes
fy = f v+0
v + vS = 970.6 ≈ 9.7 × 102 Hz . (b) The frequency heard by an observer at the cliﬀ (and thus the frequency of the sound reﬂected by
the cliﬀ, ultimately reaching your ears at some distance from the cliﬀ) is
fc = f v+0
v − vS = 1031.3 ≈ 1.03 × 102 Hz . (c) The beat frequency is fc − fy = 61 beats/s (which, due to speciﬁc features of the human ear, is too
large to be perceptible). ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08