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?

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nghodlokl · Accepted Answer

The beat frequency of two waves is:  fbeat=|f1−f2|  By rewriting this equation we get (higher frequency, so we use the plus sign):  f2=f1+fbeat=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 · Accepted Answer

Answer:   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 · Accepted Answer

Answer: 527 Hz Solution: As per the question: Beat frequency of the player, △f=4beats/s Frequency of the tuning fork, f=523Hz Now, the initial frequency can be calculated as: △f=f−fi fi=f±△f when, fi=f+△f=523+4=527Hz when, fi=f−△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.

A flute player hears four beats per second when she

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