A particle is perfroming simple harmoic motion along ` x-"axis"` with amplitude 4 cm and time period `1.2 sec`.The minimum time taken by the particle to move from x=2 cm to `x= + 4 cm` and back again is given by

Options

A. 0.6 s
B. 0.4 s
C. 0.3 s
D. 0.2 s

Answer

B

Explanation

`y=asinomegat`
Time taken by the particle to mvoe from x=0 to x=2 cm
`2=4sin((2pi)/(T))t implies t=(pi)/(6)xx(T)/(2pi)=(1.2)/(12)=0.1 sec`
and time taken by particle to move from x=0 to
`x=4` is `t_(2)=(pi)/(2)*(T)/(2pi)=(T)/(4)=(T)/(4)=0.3` sec.
Time to move from `x=2` to `x=4`
`t.=0.3-0.1=0.2sec`
Total time `=0.2xx2=0.4sec`