A particle is executing SHM. The time taken for `(3/8)^th` of oscillation from extreme positions is x. Then the time taken for the particles to complete `(5/8)^(th)` of oscillation from mean position is

Options

A. `(5x)/(4)`
B. `(7x)/(4)`
C. `(21x)/(8)`
D. `(7x)/(12)`

Answer

B

Explanation

We divide total distance 4A in 8 equal parts.

So, for a displacement of `3/8` th of and oscillation from an extreme,

Time `=T/6+T/12+T/12= (2+1+1)/(12)T= (4)/(12) T= T/3`
Given, `T/3=x or T=3x`
Now for `5/8`th of oscillation from mean

Time `=T/12 + T/6+T/6+T/12+T/12`
`rArr ` Time for `5/8` th of oscillations
`= 7/12 T= (7 xx 3x)/(12)=(7)/(4) x`