PRACTICE IT Use the worked example above to help you solve this problem. A certain piano string is supposed to vibrate at a frequency of 4.36 x 10² Hz. In order to check its frequency, a tuning fork known to vibrate at a frequency of 4.36 x 10² Hz is sounded at the same time the piano key is struck, and a beat frequency of 8 beats per second is heard. (a) Find the two possible frequencies at which the string could be vibrating. flower = fhigher = Hz Hz (b) Suppose the piano tuner runs toward the piano, holding the vibrating tuning fork while his assistant plays the note, which is at 428 Hz. At his maximum speed, the piano tuner notices the beat frequency drops from 8 Hz to 4 Hz (without going through a beat frequency of zero). How fast is he moving? Use a speed of sound of 343 m/s. m/s (c) While the piano tuner is running, what beat frequency is observed by the assistant? [Note:
PRACTICE IT Use the worked example above to help you solve this problem. A certain piano string is supposed to vibrate at a frequency of 4.36 x 10² Hz. In order to check its frequency, a tuning fork known to vibrate at a frequency of 4.36 x 10² Hz is sounded at the same time the piano key is struck, and a beat frequency of 8 beats per second is heard. (a) Find the two possible frequencies at which the string could be vibrating. flower = fhigher = Hz Hz (b) Suppose the piano tuner runs toward the piano, holding the vibrating tuning fork while his assistant plays the note, which is at 428 Hz. At his maximum speed, the piano tuner notices the beat frequency drops from 8 Hz to 4 Hz (without going through a beat frequency of zero). How fast is he moving? Use a speed of sound of 343 m/s. m/s (c) While the piano tuner is running, what beat frequency is observed by the assistant? [Note:
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:SOLUTION
(A) Find the two possible frequencies.
Case 1: f₂f₁ is already positive, so
just drop the absolute-value signs.
Case 2: f₂-f₁ is negative, so drop the
absolute-value signs, but apply an
overall negative sign.
fo
= 438 Hz.
fb = f₂-f₁4 Hz = f₂ - 4.40 x 10² Hz
f₂ = 444Hz
fb = -(f₂-f₁) · → 4 Hz = -(f₂ - 4.40 × 10² Hz)
£₂
(B) Find the speed of the observer if running toward the piano results in a beat frequency of 2 Hz.
Apply the Doppler shift to the case
where frequency of the piano string
heard by the running observer is
Compute the beat frequency:
fo
=
VO
=
436Hz
438 Hz = (436 Hz)
=
v + vo
fs V - VS
(C) What beat frequency does the assistant observe?
Apply the doppler shift equation. Now
the source is the tuning fork, so
fs = 4.40 x 10² Hz.
Hz
(436 HZ) (343 m/s) - 343 m/s = 1.57 m/s
fo=fs
v vo
(1+10)
-
343 m/s + vo
343 m/s
343 m/s
=
= (4.40 × 10² Hz) ( 343 m/s - 1.57 m/s)
fb = f₂-f1= = 442 Hz
-
436 Hz = 6 Hz
= 442 Hz
![PRACTICE IT
Use the worked example above to help you solve this problem. A certain piano string is supposed to
vibrate at a frequency of 4.36 x 10² Hz. In order to check its frequency, a tuning fork known to vibrate at
a frequency of 4.36 x 10² Hz is sounded at the same time the piano key is struck, and a beat frequency of
8 beats per second is heard.
(a) Find the two possible frequencies at which the string could be vibrating.
flower
fhigher
=
=
(b) Suppose the piano tuner runs toward the piano, holding the vibrating tuning fork while his
assistant plays the note, which is at 428 Hz. At his maximum speed, the piano tuner notices the
beat frequency drops from 8 Hz to 4 Hz (without going through a beat frequency zero). How
fast is he moving? Use a speed of sound of 343 m/s.
m/s
Hz
Hz
(c) While the piano tuner is running, what beat frequency is observed by the assistant? [Note:
Assume all numbers are accurate to two decimal places, necessary for this last calculation.]
Hz
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
The assistant adjusts the tension in the same piano string, and a beat frequency of 4.00 Hz is heard when
the note and the tuning fork are struck at the same time.
(a) Find the two possible frequencies of the string.
flower
fhigher
=
Hz
Hz
Hz
(b) Assume the actual string frequency is the higher frequency. If the piano tuner runs away from
the piano at 4.42 m/s while holding the vibrating tuning fork, what beat frequency does he hear?
f =
Hz
(c) What beat frequency does the assistant on the bench hear? Use 343 m/s for the speed of
sound.
f =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5630fc0f-1d39-4405-9289-17dd8428ca98%2Fb1f0bede-8f1e-4099-a36c-293e7903a238%2Fp92kxee_processed.png&w=3840&q=75)
Transcribed Image Text:PRACTICE IT
Use the worked example above to help you solve this problem. A certain piano string is supposed to
vibrate at a frequency of 4.36 x 10² Hz. In order to check its frequency, a tuning fork known to vibrate at
a frequency of 4.36 x 10² Hz is sounded at the same time the piano key is struck, and a beat frequency of
8 beats per second is heard.
(a) Find the two possible frequencies at which the string could be vibrating.
flower
fhigher
=
=
(b) Suppose the piano tuner runs toward the piano, holding the vibrating tuning fork while his
assistant plays the note, which is at 428 Hz. At his maximum speed, the piano tuner notices the
beat frequency drops from 8 Hz to 4 Hz (without going through a beat frequency zero). How
fast is he moving? Use a speed of sound of 343 m/s.
m/s
Hz
Hz
(c) While the piano tuner is running, what beat frequency is observed by the assistant? [Note:
Assume all numbers are accurate to two decimal places, necessary for this last calculation.]
Hz
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
The assistant adjusts the tension in the same piano string, and a beat frequency of 4.00 Hz is heard when
the note and the tuning fork are struck at the same time.
(a) Find the two possible frequencies of the string.
flower
fhigher
=
Hz
Hz
Hz
(b) Assume the actual string frequency is the higher frequency. If the piano tuner runs away from
the piano at 4.42 m/s while holding the vibrating tuning fork, what beat frequency does he hear?
f =
Hz
(c) What beat frequency does the assistant on the bench hear? Use 343 m/s for the speed of
sound.
f =
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