When a body of mass `1.0 kg` is suspended from a certain light spring hanging vertically, its length increases by `5cm`. By suspending `2.0kg` block to the spring and if the block is pulled through `10cm` and released, the maximum velocity of it in `m//s is (g = 10 m//s^(2))`

Options

A. 0.5
B. 1
C. 2
D. 4

Answer

B

Explanation

`(V_(P))/(V_(Q))=(a omega_(P))/(a omega_(Q))=(T_(Q))/(T_(P))=(6)/(3)=(2)/(1)`
`omega = sqrt((k)/(m))=sqrt((200)/(2))=10`
`omega = sqrt((k)/(m)) = sqrt((200)/(2)) = 10,V = A omega = 1 ms^(-1)`