The acceleration-displacement graph of a particle executing SHM is as shown in the figure. The velocity of the particle at y = 0 is (take `tan 14^(@)` = 0.25)

Options

A. 1 `ms^(-1)`
B. 2 `ms^(-1)`
C. 4 `ms^(-1)`
D. 8 `ms^(-1)`

Answer

C

Explanation

`a = - omega^(2) y`
Graph between .a. and .y. is a straight line
Slope of the straight line
`= omega^(2) = (1)/(tan 14^(@)) rArr (1)/(0.25) = 4`
`4 = (a)/(y), r = (8)/(A) rArr A = 2m`
`V_(max) = A omega = 2 xx 2 = 4 ms^(-1)`