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

On comparing this displacement equation with Eq. (15.2).
`y(x,t) = a sin (kx - omega t)`.
we find
(a) the amplitude of the wave is `0.005 m = 5 mm`.
(b) the angular wave number k and angular frequency `omega` are
`k = 80.0 m^(-1)` and `omega = 3.0 s^(-1)`
We, then relate the wavelength `lamda` to k through Eg. (15.6).
`lamda = 2 pi//k`
`= (2 pi)/(80.0 m^(-1))`
`= 7.85 cm`
(c ) Now, we relate T to `omega` by the relation
`T = 2 pi//omega`
`= (2 pi)/(3.0 s^(-1))`
`= 2.09 s`
and frequency , `v = 1//T = 0.48 Hz`
The displacement y at x = 30.0 cm and time t = 20 s is given by
`y = (0.0005 m) sin (80.0 xx 0.3 - 3.0 xx 2.0)`
`= (0.005 m) sin (-36 + 12 pi)`
`= (0.005 m) sin (1.699)`
`= (0.005 m) sin (97^(@)) ~= 5 mm`