# 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

148 Points
14 years ago

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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3 years ago
Dear student,

SHM equation is given as
x = 4sin(2π/T *t)
where T is time period
first find out time when particle will be at x = 2 cm
2 = 4sin(2π/T *t1)
or    sin(2π/T *t1) =1/2
or     2π/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

Hope it helps.
Thanks and regards,
Kushagra