Here, `k=1200Nm^(-1),m=3.0kg, a=2.0cm=0.2m`

(a) Frequency, `v=(1)/(T)=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx3.14)sqrt((1200)/(3))=3.2s^(-1)`

(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)=(1200xx0.02)/(3)=8ms^(-2)`

(c) Max. speed ofo the mass will be when it is passing through the mean position, which is given by

`V_(max)=aomega=asqrt((k)/(m))=0.02xxsqrt((1200)/(3))=0.4ms^(-1)`