Equation of a transverse wave travelling in a rope is given by
`y=5sin(4.0t-0.02 x)`
where y and x are expressed in cm and time in seconds. Calculate
(a) the amplitude, frequency,velocity and wavelength of the wave.
(b) the maximum transverse speed and acceleration of a particle in the rope.

Explanation

(i). Given equation is
`y=5 sin (4.0t-0.02x)`
`Comparing this with standard equation `y=A sin (omegat-kx)`, we get
Amplitude `A= 5cm`
`omega=4.0 s^(-1)` and `k=0.02 cm^(-1)`
`:.` Frequency
`f=(omega)/(2pi)=(4.0)/(2pi)=(2)/(3.14)=0.637 s^(-1)`
Velocity
`v=(omega)/(k)=(4.0)/(0.02)=200 cm//s`
wavelength `lambda=v/f=(200)/(0.637)=314 cm`
(ii) The maxium velocity,`u_(max)=omegaA=4xx5=20 cm//s`.