The displacement y(t) = A sin (omega t + phi) | vedclass

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(B) The displacement is given by $y(t) = A \sin (\omega t + \phi)$.Given $\phi = \frac{2\pi}{3}$.At $t = 0$,the displacement is $y(0) = A \sin(\phi) =

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(B) The displacement is given by $y(t) = A \sin (\omega t + \phi)$.Given $\phi = \frac{2\pi}{3}$.At $t = 0$,the displacement is $y(0) = A \sin(\phi) = A \sin(\frac{2\pi}{3})$.Since $\sin(\frac{2\pi}{3}) = \sin(120^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$,we have $y(0) = 0.866 A$.This means at $t = 0$,the graph must show a positive displacement equal to approximately $0.87 A$. Additionally,the velocity $v(t) = \frac{dy}{dt} = A\omega \cos(\omega t + \phi)$. At $t = 0$,$v(0) = A\omega \cos(\frac{2\pi}{3}) = A\omega (-0.5) = -0.5 A\omega$. Since the velocity is negative at $t = 0$,the graph must be decreasing at $t = 0$. Graph $B$ shows a positive displacement at $t = 0$ and a negative slope,which matches these conditions.

The displacement $y(t) = A \sin (\omega t + \phi)$ of a pendulum for $\phi = \frac{2\pi}{3}$ is correctly represented by which of the following graphs?

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