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Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.

Respuesta :

Answer:

Explanation:

vA / vB = √(TA/(m/L)) / √(TB/(m/L))

The lengths are the same, so the L divides out to 1

The material is identical so the mass will be directly proportional to the cross sectional area of the string

vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))

π/4 is common so divides out to 1

vA / vB = √(TA/dA²) / √(TB/dB²)

vA / vB = √(410/0.489²) / √(809/1.27²)

vA / vB = 41.407 / 23.396

vA / vB = 1.8488961...

vA / vB = 1.85

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