A transverse wave travelling in a string produce maximum transverse velocity of `3 m//s` and maximum transverse acceleration `90 m//s^(2)` in a particle. If the velocity of wave in the string is `20 m//s`. Datermine the equation of the wave ?

Answer

`y=(10 cm)sin(30tpm(3)/(2)xx+phi)`

Explanation

Maximum particle velocity
`omegaA = 3 m//s` ….(i)
Maximum particle acceleration
`omega^(2)A=90 m//s^(2)`….(ii)
Velocity of wave `(omega)/(k) = 20 m//s`….(ii)
From equation (i),(ii) and (iii) we get `omega = 30 rad//s`
`A = 0.1 m and k = 15 m^(-1)`
`:.` Equation of waveform should be
`y = A sin (omega t +kx + phi)`
`y= (0.1m) sin[(30 rad//s)t+(1.5m^(-1))x+phi]`.