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

Question

4 years ago

13

linear density of the rope is 567 g/m. What is the amplitude of the wave? A1 =

1 answer:

Serhud [2] · Answer 1 · 2022-01-29T11:35:27+03:00

Serhud [2]4 years ago

7 0

Answer:

$A = 2.43*10^{-3} m$

Explanation:

power through string can be determined as shown in figure

$P = 2\pi ^2 HVA^2F^2$

Where

P = 57.3 W

V = 13.87 m/s

H = 567 g/m

we know that

$V = f *\lambda$ $\lambda = \frac{v}{f}$

therefore $P = 2\pi ^2 HVA^2(\frac{v}{f})^2$

$57.3 = \frac{2\pi ^2 * 0.567 *13.87^{3}* A^2}{(5.54*10^{-2})^2}$

$A = 2.43*10^{-3} m$