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

Here, `K = 1200 Nm^(-), m = 3.0 kg, a = 2.0 = 0.02m`
(i) Frequency `v = (1)/(T) = (1)/(2pi) sqrt((k)/(m)) = (1)/(2 xx 3.14) sqrt((1200)/(3)) = 3.2 s^(-1)`
(ii) 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) = 8 ms^(-2)`
(iii) Max. speed of the mass will when it is passing through mean position
`V_(max) = a omega = a sqrt((k)/(m)) = 0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)`