A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 5.00 Hz, which increases to 6.00 Hz when she tightens her violin string slightly.A) what was the frequency of her violin when she heard the 5.00 Hz beat?B) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 440 Hz beat?-tighten/loosen?

Respuesta :

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

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

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

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

[tex]5Hz = |f_2-440Hz|[/tex]

[tex]f_2 = 440 \pm 5[/tex]

[tex]f_2 = 445Hz or 435Hz[/tex]

B) The violinist must loosen the string. As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

A) 445 Hz or 435 Hz

B) On tightening the string, the frequency will increase

Given:

f1=440 Hz

[tex]f_{beat}[/tex]=5.00 Hz

To find:

Frequency of violin at 5.00 Hz; f2=?

Frequency:

It is the number of occurrences of a repeating event per unit of time. Frequency is measured in hertz (Hz) which is equal to one (event) per second.

The equation that describes the beat frequency:

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

On substituting the given values:

[tex]f_{beat}=f_2-f_1\\\\5Hz=f_2-440Hz\\\\f_2=440-5\\\\f_2=435Hz[/tex]

The value of frequency can be 445Hz or 435 Hz since the value depends upon the modulus.

B) The violinist must loosen the string. As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i.e., on tightening the string, the frequency further increases as high frequency will be produced by short strings.

Thus, on tightening the string, the frequency will increase.

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