The equation of a simple harmonic motion is `x=10sin(pi/3t-pi/12)cm`. Calculate its (i) amplitude, (ii) time period, (iii) maximum speed, (iv) maximum acceleration, (v) epoch and (vi) speed after 1 s of initiation of motion.

Explanation

`x=10sin(pi/3t-pi/12)cm`.
Comparing this equation with the equation of SHM,
`x=Asin(omegat+a)` we get,
(i) amplitude, A = 10 cm.
(ii) `omega=pi/3` , so time period, `T=(2pi)/omega=(2pixx3)/pi=6s`.
(iii) maximum speed, `omegaA=pi/3xx10=(10pi)/3cm*s^(-1)`.
(iv) maximum acceleration,
`omega^(2)A=(pi/3)^(2)xx10=(10pi^(2))/9cm*s^(-2)`
(v) epoch = `-pi/12=-15^(@)`.
(vi) displacement after 1 s of initiation of motion,
`x_(1)=10sin(pi/3*1-pi/12)=10sin(60^(@)-15^(@))`
`=10sin45^(@)=10/sqrt2=5sqrt2cm`
and speed after 1 s of initiation of motion,
`v_(1)=omegasqrt(A^(2)-x_(1)^(2))=pi/3sqrt((10)^(2)-(5sqrt2)^(2))`
`=pi/3sqrt50=(5sqrt2)/3picm*s^(-1)`.