A particle exectes `S.H.M.` along a straight line with mean position `x = 0`, period `20 s` amplitude `5 cm`. The shortest time taken by the particle to go form `x = 4 cm` to `x = -3 cm` is

Options

A. `4s`
B. `7s`
C. `5s`
D. `6s`

Answer

C

Explanation

Minimum phase difference between two position
`= 53^(@) + 37^(@) = 90^(@)`

`rArr` time taken `= (T)/(4) = (20)/(4) = 5s`
OR
`omega = (2pi)/(T) = (2pi)/(20) = (pi)/(10)`

from figure `theta _(1) + theta_(2) = 53^(@) + 37^(@) = 90^(@)` or `(pi)/(2)`
`theta = omegat rArr (pi)/(2) = (pi)/(10)t rArr t = 5 sec`