Calculating Frequency, Speed, and Tension of a Guitar String: Helpful Tips"

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The discussion focuses on calculating the tension of a guitar string based on its frequency and linear density, with a specific example of a string 0.55 meters long vibrating at 283.23 Hertz. The formula for tension involves the frequency, linear density, and wavelength, but the wavelength is not directly provided, leading to questions about how to determine it. Additionally, a related problem discusses the speed of a wave on a hanging rope, emphasizing that the tension varies with distance from the bottom of the rope. The fundamental frequency concept is highlighted, indicating that the string vibrates without nodes in between. Overall, the thread provides insights into wave mechanics and tension calculations in strings and ropes.
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1) The fundamental of a guitar string whose ends are a distance 0.55m apart has a frequency 283.23 Hertz. If the linear density of the string is 0.01 kG/meter, the tension of the string (in Newtons) is

T=f^2 * linear density * wavelength^2
f=283.23
linear density = 0.01
how do find the wavelenght? I know that wavelength = velocity/frequency but the velocity of the wave on the string is not given. How does the fact that the guitar string's ends being 0.55m apart have to do with the tension of the string?


2) A heavy uniform rope is hanging freely from one end. The speed of a wave a distance 0.67 from the bottom of the rope (in meters/sec) is

how do I do this type of problem?
 
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(1) Since this is the fundamental frequancy, the string is vibrating with no nodes in between. So,

\lambda=2L

(2) Tension at distance x from bottom is \frac{Mxg}{L} where M is the mass of the rope and L is the length.
So, divide it by the mass per unit length M/L, take the square root and you get the speed.

spacetime
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Thanks spacetime! :smile:
 
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?

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