A point performs simple harmonic oscillation | vedclass

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(B) The equation of motion is $x = A \sin(\omega t + \frac{\pi}{6})$.The velocity $v$ is the derivative of displacement with respect to time: $v = \fr

Vedclass · Accepted Answer

$\frac{T}{12}$

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

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

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