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A string, 30 cm long and having a mass of 45 g, is attached to a 810 Hz oscillator at one end. The other end of the string is fixed and the string is kept under tension. The oscillator produces a transverse wave in the string, whose amplitude is 7.0 m, and which propagates with a velocity of 76 m/s. The energy of the wave is absorbed at the fixed end. In this situation, the tension in the string, in SI units, is closest to:_______.
a. 940.
b. 900.
c. 970.
d. 870.
e. 830.

Respuesta :

Answer:

870 N

Explanation:

The expression for velocity of wave in a string is given below

v = [tex]\sqrt{\frac{T}{m} }[/tex]  , T is tension and m is mass per unit length .

m = 45 x 10⁻³ / 30 x 10⁻²

= .15 kg/m

Putting the given values in the equation

76 = [tex]\sqrt{\frac{T}{.15} }[/tex]

T = 76² x .15

= 866.4

870 N  approx.

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