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

Equation of wave in string `y = 0.1 sin (30t +- (3)/(2) x + phi)` [where `phi` is initial phase]

Explanation

Maximum particle velocity `v_(max) = omegaA = 3m//s`
Maximum particle accelertion `a_(max) = omega^(2) A = 90m//s^(2)`
`a_(max) = v_(max) omega = omega xx 3 = 90 m//s^(2) rArr omega = 30s^(-1) rArr A = (3)/(omega) = (3)/(30) = 0.1 m`
`k = (omega)/(v) = (30)/(20) = (3)/(2)` [ where `v` is velocity of wave]
Equation of wave in string `y = 0.1 sin (30t+-(3)/(2)xrarr)` where `phi` is initial phase]