Y = A sin($\omega$t + $\phi$0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $Y = {A \over 2}$ and it is moving along negative x-direction. Then the initial phase angle $\phi$0 will be:

Question

Y = A sin($\omega$t + $\phi$₀) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $Y = {A \over 2}$ and it is moving along negative x-direction. Then the initial phase angle $\phi$₀ will be:

${{5\pi } \over 6}$

${{\pi } \over 3}$

${{2\pi } \over 3}$

${{\pi } \over 6}$

Solution

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