Question

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.

Answer 1

527 Hz

Step-by-step explanation:

given,

flute beat = 4 beats/s

frequency of the tuning fork = 523 Hz

to find her initial frequency.

we know

`f_(beat)=|f_1-f_2|`

`4=|f_1-f_2|`

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

Answer 2

527 Hz

Solution:

As per the question:

Beat frequency of the player,
`\Delta f = 4\ beats/s`

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

`\Delta f = f - f_(i)`

`f_(i) = f \pm \Delta f`

when

`f_(i) = f + \Delta f = 523 + 4 = 527 Hz`

when

`f_(i) = f - \Delta f = 523 - 4 = 519 Hz`

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

Hence, the initial frequency must be 527 Hz