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Solution of a particle is performing simple harmonic motion along x-axis with amplitude 4cm and time period 1.2 sec . The minimum time taken by the particle to move from x=+2 cm to x=+4 cm and back again is given by?
given
T=1.2 sec.
the min time to move from x=2 to x=4 is asked i.e.,
the time taken by the particle to move form half of its amplitude to its max amplitude
let the eqn be x=Asinwt
the time taken to go from its mean position to extreme position is T/4
time taken to cover half of its amplitude be T1
at t=0,x=0,v=Aw
at x=A/2,the time T1 is
A/2=AsinwT1,wT1=pi/6,T1=pi/6w=T/12
so time taken to move from A/2 to A is
t=T/4-T/12=T/6=1.2/6=0.2 seconds
at extreme position the velocity is zero,acceleration negative
at mid position x=A/2,
we can work it out as above and get t=T/12,t=1.2/12=.1
total time=.2+.1=.3 seconds
basically the question is time taken for going from A/2 to A and come back where A is amplitude, which comes out to be T/3 where T is time period so ans is 1.2/3 = 0.4sec
PLEASE APPROVE.
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