A particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by;

$x = a\,\sin \,\left( {\omega t + \pi /6} \right)$

After the elapse of what fraction of the time period the velocity of the particle will be equal to half of its maximum velocity?

Question

A particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by;

$x = a\,\sin \,\left( {\omega t + \pi /6} \right)$

After the elapse of what fraction of the time period the velocity of the particle will be equal to half of its maximum velocity?

Diffcult

Download our app for free and get started

Solution

$x=a \sin \left(\omega t+\frac{\pi}{6}\right)$

$\frac{d x}{d t}=a \omega \cos (\omega t+\pi / 6)$

Max. velocity $=a \omega$

$\therefore \frac{a \omega}{2}=a \omega \cos (\omega t+\pi / 6)$

$\therefore \cos (\omega t+\pi / 6)=\frac{1}{2}$

$60^{\circ}$ or $\frac{2 \pi}{6}$ radian, $=\frac{2 \pi}{\mathrm{T}} \cdot \mathrm{t}+\pi / 6=\frac{1}{2}=\pi$

$\frac{2 \pi}{\mathrm{T}} \cdot \mathrm{t}=\frac{2 \pi}{6}-\frac{\pi}{6}=+\frac{\pi}{6}$

$\therefore \mathrm{t}=+\frac{\pi}{6} \times \frac{\mathrm{T}}{2 \pi}=\left|+\frac{\mathrm{T}}{12}\right|$

Download our app
and get started for free