For the wave discribed in Exercise 15.8, plot the displacement (y) versus (t) graphs for `x=0,` 2 and 4 cm. What are the shapes of these graphs ? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ?

Explanation

Given wave equation is
`y = 3 sin (36 t +0.018 x + (pi)/(4)) cm " "…(1)`
`implies` Phase of a particle at distance x from origin of wave at time t is
`theta = 36 t + 0.018x + (pi)/(4) rad " "…(2)`
Now we are given three particles whose displacement are to be found out at different instnats of time. They are located at `x=0,` x=2 cm and x=4 cm respectively. Let us consider their simple harmonic oscillations one by one.
For particle at `x=0,` from equation (1),
`y =3 sin (36 t + (pi)/(4)) " "...(3)`
`therefore y = a sin ((2pi )/(T) t + (pi)/(4))`
(Where `(2pi)/(T) = omega = 36 (rad)/(s)) ...(4)`
From above equation we can find out vaues of displacements y at `t =0, (T)/(8), (2T)/(8), (3T)/(8), (4T)/(8),...,`
`(8T)/(8) = T` as flollows.
`(1) At t =0, y = a sin "" (pi)/(4) = (a)/(sqrt2)`
(2)` At t = (T)/(8), y = a sin ((2pi)/(T) xx (T)/(8) + (pi )/(4))`
`=a sin ((pi)/(2)) =a`
`(3) At t = (2T)/(8), y =a sin ((2pi)/(T) xx (2T)/(8) + (pi)/(4))`
`=a sin ((pi)/(2) +(pi)/(4)) =a cos ""(pi)/(4) = (a)/(sqrt2)`
`(4) At t= (3T)/(8), y= a sin ((2pi)/(T) xx (3T)/(8) + (pi)/(4))`
`=a sin (pi) =0`
`(5) At t = (4T)/(8), y = a sin ((2pi)/(T) xx (4T)/(8) + (pi)/(4))`
`= a sin (pi + (pi)/(4)) =a (-sin "" (pi)/(4)) =- (a)/(sqrt2)`
`(6)At t = (5t)/(8), y = a sin ((2pi)/(T) xx (5T)/(8) + (pi)/(4))`
`= a sin ((3pi)/(2)) =-a `
`(7) At t = (6T)/(t), y = a sin ((2pi )/(T) xx (6T)/(8) + (pi)/(4))`
`= a sin ((7pi)/(4)) =a sin (2pi - (pi)/(4)) =- (a)/(sqrt2)`
`(8) At t = (7T)/(8), y = a sin ((2pi)/(T) xx (7T)/(8) + (pi)/(4))`
`= a sin (2pi) =0`
`(9) A t t = (8T)/(8) =T,y = a sin ((2pi )/(T) xx T + (pi)/(4))`
`=a sin ((pi)/(4)) = (a)/(sqrt2) `
From above calculations, graph of `y to t` can be plotted as follows :

At `t =0,` initial phase of a particle at distance x is [from equation (2)],
`theta = 0.018 x + (pi)/(4)`
`implies ` For first particle,
`x = 0 implies theta _(1) = (pi)/(4) rad`
For second particle,
`x =4 cm implies theta_(3) =0.072 + (pi)/(4) rad`
For all of above particles wa can draw graph of `y to t ` as shown earlier. For all of them, amplitude, frequency and wavelength are same, but their initial phase are different.