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A wave is sent along the first rope transmitting a power of 57.3 W. It has a wavelength of 5.54 cm and velocity of 13.87 m/s The linear density of the rope is 567 g/m. What is the amplitude of the wave? A1 =

Respuesta :

Answer:

[tex]A = 2.43*10^{-3} m[/tex]

Explanation:

power through string can be determined as shown in figure

[tex]P  = 2\pi ^2 HVA^2F^2[/tex]

Where

P = 57.3 W

V = 13.87 m/s

H = 567 g/m

we know that

[tex]V = f *\lambda[/tex][tex]\lambda = \frac{v}{f}[/tex]

therefore [tex]P  = 2\pi ^2 HVA^2(\frac{v}{f})^2[/tex]

[tex]57.3 = \frac{2\pi ^2 * 0.567 *13.87^{3}* A^2}{(5.54*10^{-2})^2}[/tex]

[tex]A = 2.43*10^{-3} m[/tex]

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