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2 years ago

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A uniform cylindrical steel wire (density: 7.8 x 103 kg/m3), 58.0 cm long and 1.34 mm in diameter, is fixed at both ends. To wha

t tension must it be adjusted so that , when vibrating in its first overtone, it produces the note D-sharp of frequency 311 Hz? Assume that it stretches an insignificant amount. Give an answer in N. Pay attention to the number of significant figures.

1 answer:

lbvjy [14]2 years ago

3 0

Answer:

T= 354.65 N

Explanation:

Given that

ρ= 7800 kg/m³

L= 58 cm

d=1.34 mm

f= 311 Hz

T= Tension

Speed of wave ,V

V = f λ

V = f L

V= 311 x 0.58 m/s

V=180.38 m/s

Area of cross sectional

A= πr² mm²

A= 3.14 x 0.67² mm²

A=1.41 mm²

Mass = Density x Volume

$m=7800\times 1.41\times 10^{-6}\ kg/m$

m=0.0109 kg/m

Tension ,T

$T=m V^2$

T= 0.0109 x 180.38² N

T= 354.65 N

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