Question

A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and `90 m//s^(2)` respectively. Velocity of the wave is `20 m//s`. Find the waveform.

Answer 1

Correct Answer - `y=(10 cm)sin(30tpm(3)/(2)xx+phi)`
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]`.