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

Click For Summary
The discussion focuses on the effects of water leakage on the resonant frequency of a vibrating guitar string. Initially, the string's tension and mass are calculated to determine the wave speed and frequency. As water drains from the bucket at a steady rate, the tension in the string decreases, affecting the wave speed and frequency. The participants highlight the importance of understanding how the string's properties change as the water level drops, leading to an increase in frequency despite the loss of mass. The conversation emphasizes the need for clarity in calculations and reasoning regarding the physical principles involved.
chrisakatibs
Messages
2
Reaction score
0

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
 
Physics news on Phys.org
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 26 ·
Replies
26
Views
7K
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
10K