[Waves] Standing waves problem (possibly )

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SUMMARY

The discussion centers on calculating the mass of a piano string (A-string, 440Hz) that is 38.9 cm long and under a tension of 667 N. The relevant equations include the wave speed equation \( v = \sqrt{F/\mu} \) and the relationship between wavelength and length for fundamental harmonics. The correct calculation shows that the mass of the string is 0.0022 kg, derived from the linear mass density \( \mu = 0.00569 \) kg/m. A critical correction was made regarding the interpretation of the wave speed equation.

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  • Understanding of wave mechanics, specifically standing waves
  • Familiarity with the concepts of tension and mass density in strings
  • Knowledge of fundamental harmonics in vibrating strings
  • Proficiency in algebraic manipulation of equations
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  • Study the relationship between tension, mass density, and wave speed in strings
  • Learn about harmonic frequencies and their calculations in string instruments
  • Explore the effects of string length and tension on pitch in musical instruments
  • Investigate the application of wave equations in different mediums beyond strings
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Students studying physics, particularly in acoustics and wave mechanics, as well as musicians and instrument makers interested in the physics of stringed instruments.

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Homework Statement


The A-string (440Hz) on a piano is 38.9cm long and is clamped tightly at both ends. If the string is under 667-N tension, what is its mass?

Homework Equations


\lambda = vT
\mu = mass/length
v = \sqrt{F/\mu}

The Attempt at a Solution


I don't really know which equations to useI don't know if it's right
For fundamental harmonics, L = \lambda/2
so 0.389m = \lambda/2
\lambda = 0.778m
\lambda = vT
0.778 = \sqrt{F/\mu} (1/440Hz)
0.778 = \sqrt{667/\mu} (1/440Hz)
\mu = 0.00569 = mass / 0.389m
mass = 0.0022kg
 
Last edited:
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Your third equation is incorrect. The square root of F/μ is the speed of propagation v, not the wavelength.
 
oh right. correction!
 

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