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The portion of a cello string between the bridge and upper end of the fingerboard (that part of the string that is free to vibrate) is of length 60.0 cm, and this length of the string has mass 2.00 g. The string sounds an A4 note (440 Hz) when played.
Where must the cellist put a finger (what distance x from bridge) to play a D5 note (587 Hz)? For both notes, the string vibrates in its fundamental mode.

Respuesta :

Answer:

Explanation:

length of vibration l = .6 m

mass per unit length m = 2 x 10⁻³ / .6

= 3.33 x 10⁻³ kg/ m

n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]

n is frequency of vibration , l is length , T is tension in the string .

Apply this formula in the first case

440 = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

Apply this formula for second case

n = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

587 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

Dividing

[tex]\frac{440}{587}[/tex] = [tex]\frac{l}{.6}[/tex]

l = .45 m .

onyebuchinnaji

The distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.

The given parameters;

  • length o the string at A4, l = 60 cm
  • mass, m = 2.0 g
  • frequency of A4 note, f = 440 Hz
  • frequency of D5 note, f = 5587 Hz

The frequency of a sound wave in a stretched string is calculated as;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

  • T is the tension in the string
  • μ is the mass per unit length

[tex]f_1(2l_1) = f_2(2l_2)\\\\f_1l_1 = f_2l_2\\\\l_2 = \frac{f_1l_1}{f_2} \\\\l_2 = \frac{440 \times 0.6}{587} \\\\l_2 = 0.45 \ m\\\\l_2 = 45 \ cm[/tex]

Thus, the distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.

Learn more about tension in a musical string here: https://brainly.com/question/14086522

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