Given the equetion for a wave on a string
`y=0.03 sin(3x-2t)`
where y and x are in meters and t is in seconds.
(a). At t=0, what are the values of the displacement at x=0,0.1 m,0.2 m, and 0.3m ?
(b). At x=0.1 m what are the values of the displacement at t=0,0.1 s, and 0.2 s?
(c ) what is the equetion for the velocity of oscillation of the particles of the string?
(d). what is the maximum velocity of oscillation?
(e). what is the velocity of propagation of the wave?

Explanation

Given that `y = 0.03 sin (3x - 2t)`
At = ` t = 0, y = 0.03sin 3x`
When `x = 0, y = 0.03 sin 0 = 0`
When` x = 0.1, y = 0.03 sin (0.3 xx (180)/(pi)) or y = 0.03 sin (54/pi)`
`= 0.03 xx 0.2954` `x = 0.1 m, y = 0.03 sin (0.3 - 2t)`
When `t = 0, y = 0.03 sin (0.3)`
` = 8.86 xx 10^(-3) m `
When `t = 0.2, y = 0.03 sin (0.3 - 0.4)
= - 0.03 sin 0.1
= -0.03 xx sin (18/pi)
= -0.03 xx 0.0999
` = -2.997 xx 10^(-3) m `
Velocity ofparticle
`v_(p) = (dy)/(dt)`
`= -2 xx 0.03 cos (3x - 2t)`
= - 0.06 cos (3x - 2t)
This is maximum when cos `(3 x - 2) = 1`
or maximum velocity`=0.06m/s` = `6xx10^(-2) m/s `
Weknow that the equation of a progressive wave is given
`y = a sin 2 pi((x)/(lambda) - (t)/(T))`
Comparing thegiven equation, with this equation, we have
`(2pi)/(lambda) = 3` and `(2p)/(T) = 2`
or `lambda = (2p)/(3)` and `2pi n = 2` or `n - (1)/(pi)`
Now `v = nlambda = (1)/(pi) xx (2pi)/(3)` = 0.667 m/s`.