Question

A particle is executing a motion in which its displacement as a function of time is given by
`x = 3 sin (5 pi t + pi//3) + cos (5pi + pi//3)` where `x` is in `m` and `t` is in `s`. Then the motion is
A. simple harmonic with time period `0.2s`
B. simple harmonic with time period `0.4s`
C. simple harmonic with amplitude`3m`
D. not of simple harmonic but a periodic motion

Answer 1

Correct Answer - B
`x= 3 sin (5 pi t + pi//3) + cos (5 pi t + pi//3)`
`Amplitude = x_(max) = sqrt(3^(2) + 1^(2)) = sqrt(10), omega = 5 pi`
`T = (2pi)/(omega) = (2pi)/(5pi) = 0.4s`