Calculating Wave Speeds of Steel Guitar Strings

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To find the ratio of wave speeds (VA/VB) for two steel guitar strings with different diameters and tensions, the relevant equation is v = √(T/μ), where T is tension and μ is mass per unit length. String A has a tension of 410.0 N and a diameter of 0.50 mm, while String B has a tension of 820.0 N and a diameter of 1.0 mm. The mass per unit length is affected by the diameter, with String B having a greater mass per unit length due to its larger diameter. By substituting the values into the equation, the ratio of wave speeds can be calculated. Understanding these relationships is crucial for solving the problem effectively.
Ignis Radiis
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Homework Statement


Two steel guitar strings have the same length. String A has a diameter of 0.50 mm and is under 410.0 N of tension. String B has a diameter of 1.0 mm and is under a tension of 820.0 N. Find the ration of the wave speeds, VA / VB
in these two strings.

Homework Equations


V= ƒ λ
FT = FG

The Attempt at a Solution


I'm lost on where to start...:oldcry::oldconfused::oldconfused::oldcry:
 
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Ignis Radiis said:

Homework Statement


Two steel guitar strings have the same length. String A has a diameter of 0.50 mm and is under 410.0 N of tension. String B has a diameter of 1.0 mm and is under a tension of 820.0 N. Find the ration of the wave speeds, VA / VB
in these two strings.

Homework Equations


V= ƒ λ
FT = FG

The Attempt at a Solution


I'm lost on where to start...:oldcry::oldconfused::oldconfused::oldcry:
Amongst your standard equations, you should have one that relates wave speed in a string/wire to tension and mass per unit length.
 
Ignis Radiis said:

Homework Statement


Two steel guitar strings have the same length. String A has a diameter of 0.50 mm and is under 410.0 N of tension. String B has a diameter of 1.0 mm and is under a tension of 820.0 N. Find the ration of the wave speeds, VA / VB
in these two strings.

Homework Equations


V= ƒ λ
FT = FG

The Attempt at a Solution


I'm lost on where to start...:oldcry::oldconfused::oldconfused::oldcry:
v=√(T/¢)
T= tension in wire
¢= mass per unit length
 
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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