Question

A wave travelling along a string is described by, y(x,t) = 0.005 sin ( 80.0x - 3.0 x). In which the numerical constants are in SI units ( 0.005m, 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 of the wave. Also, calaculate the displacement y of the wave at a distance x = 30.0 cm and time t =20s ?

Answer 1

On comparing this displacement equation with Eq. y ( x,t) n =a sin (kx`- omegat+phi) y(x,t)=a sin (kx-omegat)`
We find
(a) the amplitude of the wave is 0.005m=5mm.
(b) the angulare wave number k and angular frequency `omega `are k `= 80.0m^(-1)` and `omega=3.0 s^(-1)`
We then relate the wavelength `lambda` to k throughEq.
`lambda= ( 2pi )/( K )`
`= ( 2pi)/( 80.0 m^(-1))= 7.85cm`
(c ) Now we relate T to `omega ` by the relation `T = ( 2pi)/( omega)`
`= ( 2pi)/( 3.0s^(-1))`
`=2.09 s`
and frequency ,`v= ( 1)/(T ) = 0.48 Hz`
The displacement y at x= 30.0 cmand time t = 20s is given by
`y = ( 0.005m ) sin (80.0 xx0.3-3.0 xx20)`
`= (0.005m) sin ( - 36+ 12pi)`
`=( 0.005m ) sin (1.699)`
`= ( 0.005m ) sin( 97^(@)) ~= 5mm`