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?

• A. T/3
• B. T/12
• C. T/8
• D. T/6

Answer 1

The Correct Option is B

To determine the fraction of the time period after which the velocity of the point is half of its maximum velocity, we need to analyze the equation of simple harmonic motion.

The given equation of motion for simple harmonic oscillation is:

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

where \omega is the angular frequency and is related to the time period T by the equation:

\omega = \frac{2\pi}{T}

The velocity of the particle in simple harmonic motion is given by the derivative of the displacement with respect to time:

v = \frac{dx}{dt} = a \omega \cos(\omega t + \frac{\pi}{6})

The maximum velocity v_{\text{max}} is:

v_{\text{max}} = a \omega

We need the velocity to be half of its maximum velocity:

\frac{v_{\text{max}}}{2} = \frac{a \omega}{2}

Substituting the velocity:

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

Cancel a \omega from both sides:

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

The cosine of which angle is \frac{1}{2}? It is at:

\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \text{ or } \omega t + \frac{\pi}{6} = \frac{\pi}{3}

Solve for \omega t:

\omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

Convert this back in terms of time t:

\omega t = \frac{2\pi}{T} t = \frac{\pi}{6}

Solving for t, we get:

t = \frac{T}{12}

Therefore, the velocity of the point will be equal to half of its maximum velocity after the elapse of \frac{T}{12} of the time period.