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A flute player hears four beats per second when she compares her note to a 523 HzHz 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.

Respuesta :

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, [tex]\Delta f = 4\ beats/s[/tex]

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

[tex]\Delta f = f - f_{i}[/tex]

[tex]f_{i} = f \pm \Delta f[/tex]

when

[tex]f_{i} = f + \Delta f = 523 + 4 = 527 Hz[/tex]

when

[tex]f_{i} = f - \Delta f = 523 - 4 = 519 Hz[/tex]

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

Answer:

527 Hz

Explanation:

given,

flute beat = 4 beats/s

frequency of the tuning fork = 523 Hz

to find her initial frequency.

we know

[tex]f_{beat}=|f_1-f_2|[/tex]

[tex]4=|f_1-f_2|[/tex]

  f₁ - f₂ = ±4

  f₁ = 523 ± 4 Hz

now,

f₁ = 527 Hz    (or)  519 Hz

when flute length increase the frequency decreases.

hence, initial frequency is equal to 527 Hz  

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