Wave on a string tied to a weight

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SUMMARY

The discussion centers on calculating the time it takes for a wave to travel along a 17 m string with a total weight of 13 kg tied to a ceiling and a 13 kg weight at the lower end. The relevant equation for wave speed is given as v = (T/μ)^(1/2), where T is the tension and μ is the linear mass density. Participants clarify that the tension T at a point y is equal to the weight of the mass below that point, leading to the expression V(y) = (T/μ)^(1/2). The solution requires rearranging and integrating to determine the total time for the wave to traverse the string.

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  • Understanding of wave mechanics and wave speed equations
  • Familiarity with tension and linear mass density concepts
  • Basic calculus for integration and rearranging equations
  • Knowledge of gravitational acceleration (g = 10 m/s²)
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  • Study the derivation of wave speed in strings using the formula v = (T/μ)^(1/2)
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Issacros
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Homework Statement


A 17 m string of weight 13kg is tied to the ceiling at its upper end, and the lower is tied to a weight of 13 kg. how long will it take a wave to get from one end to the next? (g=10 m/s^2)


Homework Equations


v = (T/u)^(1/2)


The Attempt at a Solution


well, i think mass(y)=(13+13y/17)kg
therefor T=ma= (13+13y/17)kg*19m/(s^2)
V(y)=(T/u)^(1/2)=>> (10y)^1/2

beyond this point I'm stuck. Thanks in advance.
 
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Issacros said:
well, i think mass(y)=(13+13y/17)kg
OK, this is the total mass for all points below y (measured from the bottom of the string).
therefor T=ma= (13+13y/17)kg*19m/(s^2)
Where did you get 19 m/s^2? The tension at point y is equal to the weight of the mass below that point.
V(y)=(T/u)^(1/2)=>> (10y)^1/2
Not sure where this last expression came from. Use the correct expression for T(y).

Hint:
V(y) = (T/μ)^(1/2)
dy/dt = (T/μ)^(1/2)

Rearrange and integrate to find the total time.
 

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