A particle is performing shm along x axis with amplitude 4 cm and time period 1.2sec the min time taken by particle to move from x=2 cm to x=4cm and back again is given by

Dear Pallavi Bhardwaj

SHM equation is given as

x=4sin(2∏/T *t)

 wher T is time period

first find out time when particle will be at x=2 cm

2=4sin(2∏/T *t₁)

or    sin(2∏/T *t₁) =1/2

or     2∏/T *t_{1     =}∏/6

            t1    = T/12

now find out time when particle will be at 4 cm

you can say that it will be t2=T/4 (because 4 is at extreme position)

so time taken to reach from 2cm to 4cm is  =t2- t1

                                                               =T/4 -T/12

                                                               =T/6

so total  time taken is =2*T/6

                                =T/3