A particle suspended from a vertical spring oscillates 10 times per second. 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 stretched by 0.20 cm. Take `g=pi^2ms^-2`

Explanation

a. The mea position of the particle durig vertical oscillsations is `mg/k` distance away from its position when the sprig is unstretched. At the highest point i.e. at an extreme positon, the spring is unstretched.

hence the amplitude is
`A=(mg)/k`..............i
The angular frequency is
`omega=sqrt(k/m)=2piv=(20pi)s^-1`........ii
or `m/k=1/(40pi^2)s^2`
Putting in i the amplitude is
`A=(1/(400pi^2)s^2)(pi^2m/s^2)`
`=1/400 m=0.25cm`
The maximum speed `=Aomega`
k `=(0.25cm)(20pis^-1)=5picms^-1`
b. When the spring is stretched by 0.20 cm the block is 0.25 cm-0.20 cm=0.05 cm above the meann position. The speed at this positioin will be
`v=omegasqrt(A^2-x^2)`
`=(20pis^-1)(sqrt((0.25cm)^2-(0.05cm)^2)`
`=15.4cms^-1`