A particle is subjected to two simple harmonic motions
`x_1=A_1 sinomegat`
`and x_2=A_2sin(omegat+pi/3)`
Find a the displacement at t=0, b. the maxmum speed of the particle and c. the maximum acceleration of the particle

Options

A. `a_(max)=omega^2sqrt(A_1^2+A_2^2+A_1A_2)`
B. `a_(max)=omega^2sqrt(A_1A_2)`
C. `A=A_1+A_2`
D. `A=sqrt(A_1^2+A_2^2+A_1A_2)`

Answer

A, D

Explanation

The resultant of two motions is simple harmonic of 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)`
Maximum acceleration `=omega^2A=omega^2sqrt(A_1^2+A_2^2+A_1A_2)`.