Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A point 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 the velocity of the point will be equal to half of its maximum velocity?

T/8

T/6

T/3

T/12

Step-by-Step Solution

Identify Velocity Equation: The displacement is $x = a \sin(\omega t + \pi/6)$ . Velocity $v$ is the time derivative of displacement ( $v = dx/dt$ ). $v = \frac{d}{dt}[a \sin(\omega t + \pi/6)] = a\omega \cos(\omega t + \pi/6)$
Identify Maximum Velocity: The maximum value of $\cos(\theta)$ is 1. Therefore, $v_{max} = a\omega$ .
Set Condition: We need the time $t$ when $v = \frac{v_{max}}{2}$ . $a\omega \cos(\omega t + \pi/6) = \frac{a\omega}{2}$ $\cos(\omega t + \pi/6) = \frac{1}{2}$
Solve for Time: The general solution for $\cos \theta = 1/2$ is $\theta = \pi/3$ (for the first positive instance). $\omega t + \frac{\pi}{6} = \frac{\pi}{3}$ $\omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ Substitute $\omega = \frac{2\pi}{T}$ : $\frac{2\pi}{T} t = \frac{\pi}{6}$ $t = \frac{T}{12}$

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