Using graphical means find an amplitude `a` of oscillations resulting from the superposition of the following oscillations of the same direction `:`
`(a)``x_(1)=3.0 cos (omegat+pi//3), x_(2)=8.0 sin (omega t+ pi//6),`
`(b)` `x_(1)=3.0 cos omegat, x_(2)=5.0 cos (omegat+pi//4), x_(3)=6.0 sin omegat.`

Explanation

We take a graph paper and choose an axis `(X-`axis) and an origin. Draw a vector of magnitude 3 inclined at an angle `(pi)/(3)` with the `X-` axis. Draw another vector of magnitude 8 inclined at an angle `-(pi)/(3)` (Since `sin (omegat+pi//6)=cos omegat-pi//3)` with the `X-`axi. The magnitude of the resultant of both these vectores (drawn from the origin) obtained using parallelogram law is the resultant, amplitude.
Clearly `R^(2)=3^(2)+8^(2)+2.3.8. cos ((2pi)/(3))=9+64-48xx(1)/(2)`
`=73-24=49`
Thus `R=7` units
(c) One can follow the same graphical method here but the result can be obtained more quickly by breaking into sines and cosines and adding `:`
Resultant `x=(3+(5)/(sqrt(2)))cos omegat +(6-(5)/(sqrt(2)))sin omegat `
`=A cos(omegat+alpha)`
Then `A^(2)=(3+(5)/(sqrt(2)))^(2)+(6-(5)/(sqrt(2)))^(2)=9+25+(30-60)/(sqrt(2))+36`
`=70-15sqrt(2)=70-21.2`
So, `A=6.985~~7 `units
Note- In using graphical method convert all oscillations to either sines or cosines but do not use both.