A particle executing simple harmonic motion has angular frequency `6.28 s^(-1)` and amplitude `10 cm`. Find `(a)` the time period, `(b)` the maximum speed, `(c)` the maximum acceleration, `(d)` the speed when the displacement is `6 cm` from the mean position, `(e)` the speed at `t = 1//6 s` assuming that the motion starts from rest at `t = 0`.

Explanation

(a) Time period `= (2pi)/(omega) = (2pi)/(6.28) s= 1s`.
(b) Maximum speed `= Aomega = (0.1 m)(6.28 s^(-1))`
(c) Maximum accelertion `= Aomega^(2)`
`= (0.1 m) (6.28 s^(-1))^(2)`
`= 4m//s^(2)`.
(d) `v = omegasqrt(A^(2) - x^(2)) = (6.28 s^(-1)) sqrt((10 cm)^(2) - (6 cm)^(2)) = 50.2 cm//s`.
(e) At `t = 0`, the velcocity is zero i.e., the particle is at an extreme. The equation for displacement may be written as
The velocity is `v = -A omega sin omegat`.
At `t = (1)/(6)s, v = -(0.1m)(6.28 s^(-1)) sin ((6.28)/(6))`
`(-0.628 m//s) sin'(pi)/(3) = 54.4 cm//s.`