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(A) The displacement equation is given by $x(t) = A \sin(\omega t + \delta)$.At $t = 0$,the position is $x = \frac{A}{2}$.Substituting these values: $

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(A) The displacement equation is given by $x(t) = A \sin(\omega t + \delta)$.At $t = 0$,the position is $x = \frac{A}{2}$.Substituting these values: $\frac{A}{2} = A \sin(\omega(0) + \delta) \Rightarrow \sin \delta = \frac{1}{2}$.This gives two possible values for $\delta$ in the range $[0, 2\pi)$: $\delta = \frac{\pi}{6}$ or $\delta = \frac{5\pi}{6}$.The velocity of the particle is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \delta)$.At $t = 0$,$v(0) = A\omega \cos \delta$.Since the particle moves along the positive $x$-axis,the velocity must be positive $(v > 0)$,which implies $\cos \delta > 0$.For $\delta = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} > 0$ (Valid).For $\delta = \frac{5\pi}{6}$,$\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} Therefore,the correct value is $\delta = \frac{\pi}{6}$.

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

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