A `6.00m` segment of a long string has a mass of `180 g`. A high-speed photograph shows that the segment contlains four complete cycles of a wave. The string is vibrating sinusoidally with a frequency of `50.0 Hz` and a peak - to - valley dispalcement of `15.0 cm`. (The "peak-to-valley" displacment is the vertical distance from the furthest positive displacement to the farther negative displacement.) (a) Write the funcation that describe this wave traveling in the positive `x` direction. (b) Determine the average power being supplied to the string.

Answer

(a) `y = (7.50 cm) sin ((4pi)/(3) xx -314t + phi)`
(b) `(2025pi^(2))/(32) = 625 W`

Explanation

(a) `f = 500Hz`
`mu = (180 xx 10^(-3))/(6) = 3 xx 10^(-2)Kg//m`
`2A = 15 xx 10^(-2)m`
`A = 7.5 xx 10^(-2)m`
`4lambda = 6 , lambda = 3//2 m`. and `v = flambda = 75 m//sec`
`omega = 2pif = 100pi` also `K = (omega)/(v) = (4pi)/(3)`
If phase constant is `phi` then
`:.` equation is `y = 7.5 xx 10^(-2) sin [100pit-(4pi)/(3)x + phi]`
(b) `P_(av) = 2pi^(2)f^(2)A^(2)muv = (2025)/(32) pi^(2) = 625 w`