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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 *t1 =∏/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
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