A particle suspended from a vetical spring oscillastes 10 times per sencon. At the highest point of oscillation the spring becomes unstretched. A. Find the maximum speed of the block. B. Find the speed when the spring is stretche by 0.20 cm. Take `g=pi^2ms^-2`

Explanation

(a) When a spring in equilibrium, its extension `=(mg)/(k)`
At highest point of block, spring is in its natural length
Hence, amplitude `A=(mg)/(k)`
`f=10Hz, omega=2pif=20pirad//s`
`omega=sqrt(k/m)impliesk/m=omega^2`
`v_(max)=Aomega=m/kgomega=(gomega)/(omega^2)=g/omega=(pi^2)/(20pi)`
`=(pi)/(20)m//s`
`A=(mg)/(k)=(g)/(omega^2)=(pi^2)/(400pi^2)=(0.25)/(100)m=0.25cm`
(b) When extension is `0.20cm`, displacement from equilibrium position `=0.25-0.20=0.05cm`
`x=0.05cm`
`v=omegasqrt(A^2-x^2)=20pisqrt((0.25)^2-(0.05)^2)`
`=2pisqrt6cm//s`
`=15.4cm//s`