Caculating the tention, max accaleration and velocity of a violin string

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SUMMARY

The discussion focuses on calculating the tension, maximum acceleration, and velocity of a violin string made of steel wire with a diameter of 0.25 mm, specifically for the E-string tuned to 660 Hz. The required tension to achieve this frequency is determined to be 73.61 N, while the maximum velocity of the string is calculated to be 429 m/s. Additionally, the maximum acceleration of the string's parts, when vibrating with an amplitude of 5 mm, is found to be 85,897 m/s².

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  • Understanding of linear mass density (μ) in physics
  • Familiarity with wave equations and harmonic motion
  • Knowledge of basic calculus for deriving formulas
  • Experience with physical properties of materials, specifically steel
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  • Study the relationship between tension and frequency in string instruments
  • Learn about wave propagation in different materials
  • Explore the effects of amplitude on maximum acceleration in harmonic motion
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Musicians, physics students, and string instrument makers who are interested in the mechanics of string vibrations and the physical principles governing sound production in violins.

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Homework Statement



String on the violin is made of steel wire with a diameter of 0.25 mm. With what force must the string be strained (stretched) so that its basic frequency will equal 660 Hz (E-string)? What is the maximum velocity and acceleration of the parts of the string, if we evoke basic vibration with amplitude of 5 mm? In the rest, the fastened string is long l = 325 mm, the density of steel is 7800 kg/m3.

Homework Equations



μ= m/L
v= 2Lf
a(max)= ω²A

The Attempt at a Solution



First part: With what force must the string be strained so that its basic frequency will equal 660 Hz (E-string)?

Volume of cylinder: V= πr²h= π(1/d)²L= 16 x 10^-9 m³, ρ= 7800 kg/m³
Mass of the string: m= Vρ= 0.00013 kg
Linear mass density: μ= m/L (kg/m)→ μ= 0.0004 kg/m

v= 2Lf= 2(0.325m)(660 hz)= 429 m/s
v= sqrt(T/ μ) → T or F= μv²
F= 73.61 N

Second part: What is the maximum velocity and acceleration of the parts of the string, if we evoke basic vibration with amplitude of 5 mm?

v= 2Lf= 2(0.325m)(660 hz)= 429 m/s
a(max)= ω²A= (2πf)²A= 85897m/s²
 
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