How Does Water Leakage Affect the Frequency of a Vibrating Guitar String?

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SUMMARY

The discussion focuses on the effects of water leakage from a bucket on the resonant frequency of a vibrating guitar string. The initial resonant frequency calculated is 50.754 Hz, with a wave speed of 143.15997 m/s. As water drains at a rate of 2.00 g/s, the wave speed changes at a rate of 0.01129 m/s², while the frequency increases over time. The fundamental wavelength is also affected, indicating that the tension in the string alters as the mass of water decreases.

PREREQUISITES
  • Understanding of wave equations and string tension
  • Knowledge of fundamental frequency calculations
  • Familiarity with mass density and its impact on wave properties
  • Basic principles of harmonic motion in strings
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Homework Statement



A 210 g bucket containing 12.5 kg of water is hanging from a steel guitar string with mass m = 8.58 g, density p = 7750 kg/m3, and diameter d = 1.00 mm. A) When the wind blows, it causes the cord to vibrate at what resonant frequency? B) Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s. At what rate are the (i) wave speed, (ii) wavelength of the fundamental mode of vibration, and (iii) frequency changing? C) How long will it take until the vibration is no longer audible (between 20 Hz and 20,000 Hz)?

Homework Equations



Many, wave equaitons, string tension etc.

The Attempt at a Solution



First parts are here to confirm, need help with last part, I believe it should be just plugging 20 Hz into an equation.

V=sqrt(T*L/mass)
T=mg=(12.5 0.210)*9.81=124.6851 N
L=volume /area =(0.00858/7750)*10^(6)/(3.14*0.5^2)=1.41031436m
a)V=sqrt(124.6851*1.41031436/0.00858)=143.15997 m/s
frequency=velocity /2L=143.15997/(2*1.41031)=50.754 Hz
B)i) dv/dt= -0.0196*sqrt(L/(Tm))=0.01968*sqrt(1.41031436/(124.6851*0.00858))=0.0225960m/s^2
wave speed change rate = 0.01129 m /s^2
ii)wavelength of the fundamental mode of vibration change=1/(2*50.754*124.6851)=0.00007901056 m/s
iii)frequency changing=0.01129 /(2*1.4103143)=0.00400265387
 
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It helps if you explain your reasoning instead of just listing a bunch of equations.
The last question concerns the rate that the resonant frequency excited by the wind will change as water leaves the bucket.

in B - you appear to have the wavelength increasing with time
I think you should check your reasoning there - what sets the fundamental wavelength on a string fixed at both ends?

You left off the units in B(iii) too - the fundamental frequency appears to be increasing in time, as does the wave speed.
What is it about the string that changes as the bucket drains?
 

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