I don\'t know where I went wrong for the second part please help. A violin strin

Question

I don't know where I went wrong for the second part please help.

A violin string has a length of 0.310 m and is tuned to concert G, with f_G = 392 Hz. How far from the end of the string must the violinist place her finger to play concert A, with f_A = 440 Hz? If this position is to remain correct to one-half the width of a finger (that is, to within 0.540 cm), what is the maximum allowable percentage change in the string tension? Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.%

Explanation / Answer

If the length of the string is L, the fundamental harmonic is the one produced by the vibration whose nodes are the two ends of the string, so L is half of the wavelength of the fundamental harmonic. Hence:
f = 1/(2L) * (T/) ........(*)
where T is the tension, is the linear mass, and L is the length of the vibrating part of the string.
Therefore, frequency is inversely proportional to length:
f1/f2 = L2/L1
Length corresponding to concert A (440Hz) is:
L2 = L1 f1/f2
L2 = 0.31 * 392/440 = 0.27618 m
"If this position is to remain correct to one-half the width of a finger" etc. - I understand this part of question as if allowed tolerance for L2 is ±0.54cm = ± 0.0054 m
That would correspond to allowed relative change of length (let's denote it by ) is:
= L / L2 = ±0.0054/0.27618 = ± 0.01955= ± 1.955%
because length is inversely proportional to square root of tension we can write
1+ = 1/(1+)
where is maximum allowable relative change of tension.
square both sides:
(1+)² = 1/(1+)
= 1/(1+)² - 1
= 1/(1± 0.01955)² - 1 = -0.03799 = -3.799%

or +0.040277 =+4.0277%

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.