The displacement of a vibrating particle at time t is given by `x=A'sin(pi/6t)+B'cos(pi/6t)`, where A' = 0.03m, B' = 0.04m. Calculate the (i) amplitude, (ii) epoch, (iii) displacement, velocity and acceleration of the particle after 2 seconds.

Explanation

`x=A.sin(pi/6t)+B.cos(pi/6t)" "...(1)`
Let `A.=AcosalphaandB.=Asinalpha`
`therefore" "A^(2)(cos^(2)alpha+sin^(2)alpha)=A.^(2)+B.^(2)`
or, `A=sqrt(A.^(2)+B.^(2))andtanalpha=(B.)/(A.)`
So, `x=Asin(pi/6t)cosalpha+Acos(pi/6t)sinalpha`
or, `x=Asin(pi/6t+alpha)" "...(2)`
From this equation we get,
(i) amplitude, `A=sqrt(A.^(2)+B.^(2))`
`=sqrt((0.03)^(2)+(0.04)^(2))=0.05m`
(ii) epoch, `alpha="tan"^(-1)(B.)/(A.)="tan"^(-1)4/3`
(iii) when t = 2 s, we get from equation (1)
displacement, `x=A.sin(pi/6t)+B.cos(pi/6t)`
`=0.03"sin"pi/3+0.04"cos"pi/3`
`=0.03xxsqrt3/2+0.04xx1/2`
`=0.04598m~~0.046m`
From equation (2) we get, `omega=pi/6"rad"*s^(-1)`
So, velocity after 2 s,
`v=omegasqrt(A^(2)-x^(2))`
`=pi/6sqrt((0.05)^(2)-(0.046)^(2))=1.03m*s^(-1)`
Again, acceleration after 2s,
`a=omega^(2)x=(pi/6)^(2)xx0.046=1.26m*s^(-2)`.