A wave travelling along a string is discribed by.
`y(x,t) = 0.005 sin (80.0 x - 3.0 t)`.
in which the numerical constants are in SI units `(0.005 m, 80.0 rad m^(-1), and 3.-0 rad s^(-1))`. Calculate (a) the amplitude, (b) the wavelength, and (c ) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and t = 20 s ?

Explanation

Given wave equatin `:y =0.005 sin (80 x - 3y)m` Comparing with standard form of wave equatin,` y = a sin (kx - omega t ) `we get,
(a) Amplitude `a = 0.005 m `
(b) Wave -vector `k = 80.0 rad//m`
`implies (2pi )/(lamda) =80 implies lamda = (2 xx 3.14)/(80) =0.0785m`
(c ) Angular frequency `omega = 3.0 rad//s`
`implies (2pi )/(T) = 3 implies T = ( 2 xx 3.14)/(3) = 2.093s`
Frequency `f = 1/T = (1)/(2.093) = 0.4778Hz`
Displacement `y =0.005 sin (80x-3t)`
`therefore y =0.005 sin (80 xx 0.3 -3 xx 20)`
`=0.005 sin (-36)`
`=-0.005 sin (36)`
`=-0.005 sin (36 xx (180)/(3.14))`
`(because 36 rad = (36 xx (180)/(3.14))^(@))`
`=-0.005 sin (2064)^(@)`
`=-0.005 sin (1800^(@) + 264^(@)) =-0.005 sin (264^(@))`
`(because 1800^(@)=5 xx360^(@)=5 xx 2pi rad =10 pi` rad is the period of sine function)
`=-0.005 sin (270^(@) -6^(@))`
`therefore y =- -0.005 (-cos 6^(@))`
`therefore y =0.005 cos 6^(@) = (0.005) (0.9945) =0.00497m`
`therefore y = 4.97mm`