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 ?

A wave travelling along a string is discribed by. y(x,t) = 0.005 sin

1.	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 ?
Answer» Solution :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`