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 particle to complete ( 5 8 )^ th of oscillation from mean position is

Step-by-step solution

We divide total distance 4 A in 8 equal parts. So, for a displacement of 3 8 th of an 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=3 x Now, for 5 8 th of oscillation from mean Time = T 12 + T 6 + T 6 + T 12 + T 12 Time for 5 8 th of oscillations = 7 12 T= 7 3 x 12 = 7 4 x