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(B) The extreme positions are $x_1 = -3 \, cm$ and $x_2 = 9 \, cm$. The mean position $x_0$ is the midpoint of the extremes: $x_0 = \frac{-3 + 9}{2} =

Vedclass · Accepted Answer

$x = 3 + 6 \sin(4\pi t + 7\pi/6)$

Vedclass · Answer

(B) The extreme positions are $x_1 = -3 \, cm$ and $x_2 = 9 \, cm$. The mean position $x_0$ is the midpoint of the extremes: $x_0 = \frac{-3 + 9}{2} = 3 \, cm$. The amplitude $A$ is the distance from the mean position to an extreme: $A = 9 - 3 = 6 \, cm$. The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \, rad/s$. The general equation for $SHM$ is $x(t) = x_0 + A \sin(\omega t + \phi)$. Substituting the values: $x(t) = 3 + 6 \sin(4\pi t + \phi)$. At $t = 0$,$x = 0$: $0 = 3 + 6 \sin(\phi) \implies \sin(\phi) = -1/2$. This gives $\phi = 7\pi/6$ or $11\pi/6$. Since the particle is moving in the negative direction at $t = 0$,the velocity $v = \frac{dx}{dt} = 6(4\pi) \cos(4\pi t + \phi)$ must be negative at $t = 0$. For $\phi = 7\pi/6$,$\cos(7\pi/6) = -\sqrt{3}/2 For $\phi = 11\pi/6$,$\cos(11\pi/6) = \sqrt{3}/2 > 0$ (Incorrect). Thus,the equation is $x = 3 + 6 \sin(4\pi t + 7\pi/6)$.

$A$ particle performs $SHM$ on the $x-$axis with a time period of $0.5 \, s,$ such that its velocity is zero at $x = -3 \, cm$ and at $x = 9 \, cm$. It was located at $x = 0$ and moving in the negative $x-$direction at $t = 0$. The equation of $SHM$ of the particle is:

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