A vertical spring-mass system with lower end of spring is fixed, made to undergo small oscillations. If the spring is stretched by `25cm`, energy stored in the spring is `5J`. Find the mass of the block if it makes `5` oscillations each second.

Answer

`(16)/(10pi^(2)) = 0.16 Kg`

Explanation

`(1)/(2)Kx^(2) = 5`
`(1)/(2) k xx ((25)/(100))^(2) = 5`
`k = 160`
`omega = 10pi = sqrt((K)/(m))`
`m = (K)/(100pi^(2))`
`m = (160)/(100pi^(2)) = (16)/(10pi^(2)) = 0.16 kg`