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

Time period = `(2pi) /(omega) = (2pi)/(6.28) s= 1 s.`
(b) Maximum speed = `A omega = (0.1m) (6.28 s^(-1))`
(c ) Maximum acceleration = `A omega^(2)`
= (0.1m) (6.28 `s^(-1))^(2) = 4 m//s^(2)`
(d) v =` omega sqrt(A^(2)-x^(2))`
=`(6.28s^(-1)) sqrt((10 cm)^(2)-(6cm^(2)))=50.2 cm//s.`
(e) At t = 0, the velocity is zero i.e., the particle is at an extreme. The equation for displacement may be written as
x = A cos `omega`t.
The velocity is v = – A `omega sin omega`t
At `t = (1)/(6) s , " " v = -(0.1m) (8.28 s^(-1)) sin ((6.28)/(6))`
= ` (-0.628 m//s) sin ""(pi)/(3)`
= - 54.4 cm/s . (towards mean position)