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Two particles execute SHM of same amplitude and same time period about same mean position but with a phase difference.At an instant they are found to cross each other at x=A/3.Then phase difference between them is

Asked by Selvabhavani25 11th March 2019, 4:20 PM
Answered by Expert
Answer:
displacement of two particles undergoing SHM with same amplitude and same period is given by
 
x1 = A sin φ1
 
x2 = A sin φ2
 
if x1 = x2 = A/3  , then  sin φ1 = sin φ2 = 1/3
 
Then φ1 = sin-1 (1/3) ≈ 19.5 ° , if φ2 is different phase angle and sin φ2 = 1/3 , then φ = (180 - 19.5 ) = 160.5° [ because sin(180-θ) = sin θ ]
 
phase difference   (φ2 - φ1 ) = 160.5 - 19.5 = 141°
Answered by Expert 11th March 2019, 5:26 PM
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