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

(i) Spring consant, `k=1200Nm^(-1)`
Mass, m=3kg
Displacement ,A=2.0cm =0.02 M
Frequency of oscillation v, is given by the relation:
`v=(1)/(T)=(1)/(2pi)sqrt((k)/(m))`
Where , T is the time period
`v=(1)/(2xx3.14)sqrt((1200)/(3))=3.18//s`
Hence,the frequency of osillation is `3.18//s`.
(ii) Maximum acceleration (a) is given by the relation
`a=omega^(2)A`
Where,
omega =Angular Frequency =`sqrt((k)/(m))`
A=Maximumdosplacement
`a=(k)/(m)A=(1200xx0.02)/(3)=8ms^(-2)`
Hence ,the maximum accelration of the mass is `8.0m//s^(2)`
(iii) Maximum velocity , `v_(max)=Aomega` Hence,the maximum velocity of the mass is `0.4m//s`.