A particle is subjected to two simple harmonic motions
`x_1=A_1 sinomegat`
`and x_2=A_2sin(omegat+pi/3)`
Find the maximum speed of the particle and the maximum acceleration of the particle

Explanation

a. At t=0, `x_1=A_1sinomegat=0`
`and x_2=A_2sin(omegat+pi/3)`
`=A+2sin(pi/3)=(A_2/sqrt3)/2`
Thus the resultant displacement at t=0 is
`x=x_1+x_2=(A_2sqrt3)/2`
b. The resultant of the two motions is a simple harmonic motion of the same angular frequency omega. The amplitude of the resultant motion is
`A=sqrt(A_1^2+A_2^2+2A_1A_2cos(pi/3))`
`=sqrt(A_1^2+A_2^2+A_1A_2)`
the maximum speed is
`=v_(max)=Aomega=omegasqrt(A_1^2+A_2^2+A_1A_2)`
c. The maximum acceleration is
`a_(ma)=Aomega^2=omega^2 sqrt(A_1^2+A_2^2+A_1A_2)`