Question

while completing it's third oscillation during linear S.H.M, a particle is at -√3/2A, heading to the mean position. Determine the phase angle .
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Answer 1

Answer:

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Answer 2

Answer:

17π/3

Explanation:

x = √3/2 A

A sin theta = √3/2 A

sin theta = √3/2

theta = sin-¹(√3/2)

theta = π+π/3 or theta =2π - π/3

since particle is in moving from negative extremity to mean position phase is in 4th quadrant

theta =2π - π/3

As particle is in 3rd oscillation it has already completed 2 oscillations

hence phase angle = 2×2π + 2π - π/3

= 4π + 2π - π/3

= 6π - π/3

= 17π /3.