contestada

The guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m by pressing on the string with a finger. What is the fundamental frequency, in hertz, of the new tone that is produced when the string is plucked?

Respuesta :

Cricetus

Answer:

The fundamental frequency is "190.52 Hz".

Explanation:

The given question is incomplete, please find attachment of the complete problem.

The given values are:

Frequency,

f = 146.8 Hz

Length of D-string,

L = 0.61

As we know,

⇒ [tex]f = \frac{v}{2L}[/tex]

On putting the given values, we get

⇒ [tex]146.8 = \frac{v}{(2\times 0.61)}[/tex]

⇒       [tex]v=146.8\times 1.22[/tex]

⇒       [tex]v=179 \ m/s[/tex]

So,

The new length will be:

⇒  [tex]f = \frac{179}{(2\times 0.47)}[/tex]

⇒     [tex]=\frac{179}{0.94}[/tex]

⇒     [tex]=190.52 \ Hz[/tex]

Ver imagen Cricetus
ajeigbeibraheem

The fundamental frequency, in hertz, of the new tone that is produced when the string is plucked is 168.84 Hz

Let assume that the oscillating length on the guitar is 0.62 m,

If the guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m.

Then the new length of the oscillating string L' = (0.62 - 0.15 )m = 0.47 m

The fundamental frequency of the new tone can be computed by using the formula:

[tex]\mathbf{f = \dfrac{1}{2L'} \sqrt{\dfrac{T}{m/L}}}[/tex]

where;

  • mass (m) is assumed to be 1.5× 10⁻³ kg, and;
  • tension T = 4Lf² m

T  = 4× 0.62× 147 × 1.5× 10⁻³  (assuming old fundamental frequency = 147)

T  = 4× 0.62× 147 × 1.5× 10⁻³

T = 80.39 N

∴

[tex]\mathbf{f = \dfrac{1}{2 \times 0.47} \sqrt{\dfrac{80.39}{1.5 \times 10^{-3}/0.47}}}[/tex]

f = 168.84 Hz

Therefore, the fundamental frequency, in hertz, of the new tone that is produced when the string is plucked is 168.84 Hz

Learn more about frequency here:

https://brainly.com/question/14316711?referrer=searchResults