A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing.
(a) the frequency of oscillation of the mass.
(b) the maximum acceleration of the mass.
(c) the maximum speed of the mass.

Explanation

`K = 1200 N m^(-1) ,m = 3 kg A = 2cm = 0.02m`
(a) Frequecny,
`f = (1)/(2pi) sqrt((K)/(m)) = (1)/(6.28) sqrt((1200)/(3)) = 3.2 Hz`
(b) Acceleration `a = omega^(2) y = (K)/(m)y` Acceleration will be maximum when `y` is maximum i.e. `y=A`
`:.` Max.acceleration,
`a_(max) = (KA)/(m) = (1200 xx 0.02)/(3) = 8ms^(-2)`
(c) Maximum speed of the mass will be when it is passing through the mean position, given by
`V_(max) = A omega =A sqrt((K)/(m)) = 0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)`