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A particle executes S.H.M. of amplitude A along x-axis. At  t = 0, the position of the particle is x=\tfrac{A}{2}  and it moves along positives x-axis. The displacement of particle in time t is x=A \sin (\omega t+\delta),then the value \delta will be

 

Option: 1

\frac{\pi}{4}


Option: 2

\frac{\pi}{2}


Option: 3

\frac{\pi}{3}


Option: 4

\frac{\pi}{6}


Answers (1)

best_answer

\begin{aligned} & \operatorname{Cos} \theta=\frac{\mathrm{A}}{2 \times \mathrm{A}}=\frac{1}{2} \\ & \theta=\frac{\pi}{3} \\ & \delta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6} \end{aligned}        

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Anam Khan

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