A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(

Question

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?

Accepted Answer

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