ajedrezlaproa6j

2022-01-05

A flute player hears four beats per second when she compares her note to a 523 Hz tuning fork (the note C). She can match the frequency of the tuning fork by pulling out the tuning joint to lengthen her flute slightly. What was her initial frequency?

nghodlokl

The beat frequency of two waves is:
${f}_{beat}=|{f}_{1}-{f}_{2}|$
By rewriting this equation we get (higher frequency, so we use the plus sign):
${f}_{2}={f}_{1}+{f}_{beat}=523+4=527Hz$
When the length of the flute is increased, its wavelength also increases thereby its frequency decreases. In order to be equal to the reference frequency (523 Hz), the frequency of the flute must be greater than the reference frequency. So we use the plug sign

Jonathan Burroughs

Initial Frequency $=527Hertz$
Explanation:
Given
Frequency of beat $=4Hertz$
Frequency of Fork $=523Hertz$
Let $Fb=$ Frequency of beat
$Ff=$ Frequency of Fork
The wave length and the length of the flute are directly proportional.
So, when the wavelength gets increased, the length of the flute is also increased.
But it is not so for the frequency.
When the wavelength and the length of the flute are increased, the frequency gets reduced.
For the flute player frequency to equate the frequency of the turning fork, the frequency of the flute must be greater than the frequency of the fork.
So, Frequency $=4+523=527Hertz$

karton

527 Hz
Solution:
As per the question:
Beat frequency of the player, $\mathrm{△}f=4beats/s$
Frequency of the tuning fork, $f=523Hz$
Now, the initial frequency can be calculated as:
$\mathrm{△}f=f-{f}_{i}$
${f}_{i}=f±\mathrm{△}f$
when, ${f}_{i}=f+\mathrm{△}f=523+4=527Hz$
when, ${f}_{i}=f-\mathrm{△}f=523-4=519Hz$
But we know that as the length of the flute increases the frequency decreases
Hence, the initial frequency must be 527Hz.

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