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frosja888 [35]
3 years ago
15

Two steel guitar strings have the same length. String A has a diameter of 0.513 mm and is under 403 N of tension. String B has a

diameter of 1.29 mm and is under a tension of 800 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.
Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer:

\frac{v_{A}}{v_{B}} = 1.785

Explanation:

T_{A} = Tension force in string A = 403 N

T_{B} = Tension force in string B = 800 N

d_{A} = diameter of string A = 0.513 mm

d_{B} = diameter of string B = 1.29 mm

v_{A} = wave speed of string A

v_{B} = wave speed of string B

Ratio of the wave speeds is given as

\frac{v_{A}}{v_{B}} = \sqrt{\frac{T_{A}}{T_{B}}} \left ( \frac{d_{B}}{d_{A}} \right )

\frac{v_{A}}{v_{B}} = \sqrt{\frac{403}{800}} \left ( \frac{1.29}{0.513} \right )

\frac{v_{A}}{v_{B}} = 1.785

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