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(B) The equation of motion is $x = a \sin(\omega t + \pi/6)$.The velocity $v$ is given by the derivative of displacement with respect to time: $v = \f

Vedclass · Accepted Answer

$T/12$

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(B) The equation of motion is $x = a \sin(\omega t + \pi/6)$.The velocity $v$ is given by the derivative of displacement with respect to time: $v = \frac{dx}{dt} = a\omega \cos(\omega t + \pi/6)$.The maximum velocity is $v_{max} = a\omega$.We are given that the velocity is half of its maximum velocity: $v = \frac{1}{2} v_{max} = \frac{a\omega}{2}$.Equating the two expressions: $a\omega \cos(\omega t + \pi/6) = \frac{a\omega}{2}$,which simplifies to $\cos(\omega t + \pi/6) = 1/2$.Since $\cos(60^{\circ}) = 1/2$ or $\cos(\pi/3) = 1/2$,we have $\omega t + \pi/6 = \pi/3$.Substituting $\omega = 2\pi/T$: $(2\pi/T)t + \pi/6 = \pi/3$.$(2\pi/T)t = \pi/3 - \pi/6 = \pi/6$.Solving for $t$: $t = (\pi/6) 	imes (T/2\pi) = T/12$.

$A$ particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period will the velocity of the particle be equal to half of its maximum velocity?

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