P18_097.PDF

97. Thesirenisbetweenyouandthecliﬀ, movingawayfromyouandtowardsthecliﬀ. Both“detectors”(you

and the cliﬀ) are stationary, so v D =0 in Eq. 18-47 (and see the discussion in the textbook immediately

after that equation regarding the selection of

(a) With f =1000 Hz, the frequency f y you hear becomes

f y = f v +0

v + v S

=970 . 6

≈

9 . 7

10 2 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

f c = f v +0

v − v S

=1031 . 3

≈

1 . 03

10 2 Hz .

f y =61 beats/s (which, due to speciﬁc features of the human ear, is too

signs). The source is the siren with v S =10 m/s. The

problem asks us to use v =330 m/s for the speed of sound.

large to be perceptible).