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Asked by adjacentcalliber10 | 14 May, 2019, 11:08: AM
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amplitude = (1/2)( maximum displacement - minimum displacement) = (1/2) [ 6 - (-2) ] = 4 cm
 
equlibrium posistion = Maximum displacement - amplitude = 6-4 = 2 cm
 
Displacement x as a function of time t for  a particle that exhbits SHM with amplitude 4 and equilibrium posistion 2 is given by,  
 
x = 4 sin(ωt+φ)+2          ............................. (1)
 
where ω is angular frequency and φ is initial phase.
 
at t0 , we have displacement x = 0 ;   hence  from eqn.(1) we get 4 sin(ωt0+φ)+2 = 0 
 
sin(ωt0+φ) = -(1/2)    or  (ωt0+φ) = π+(π/6) = (7/6)π ...............................(2)
 
at t0+1 , we have displacement x = 0; hence  from eqn.(1) we get  4 sin(ωt0+ω+φ)+2 = 0 
 
sin(ωt0+ω+φ) = -(1/2)    or  (ωt0+ω+φ) = 2π-(π/6) = (11/6)π .............(3)
 
from eqn.(2) and (3), we get ω = (2/3)π
Answered by Thiyagarajan K | 15 May, 2019, 09:18: AM

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