Test Index

BITSAT 2016 Paper

Section: Physics

© examsnet.com

Question : 36 of 150

Marks: +1, -0

A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is

8 \times 10^{-3}

Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45°.

$y = 0.1 \sin\left(\pm 4t + \; \; \frac{\pi}{4}\right)$
$y = 0.2 \sin\left(\pm 4t + \; \; \frac{\pi}{4}\right)$
$y = 0.1 \sin\left(\pm 2t + \; \; \frac{\pi}{4}\right)$
$y = 0.2 \sin\left(\pm 2t + \; \; \frac{\pi}{4}\right)$

Solution:

The displacement of a particle in S.H.M. is given by :

y = a \sin(\omega t + \phi)

velocity

= \frac{dy}{dt} = \omega a \cos(\omega t + \phi)

The velocity is maximum when the particle passes through the mean position i.e.

\left(\frac{dy}{dx}\right)_{\text{max}} = \omega a

The kinetic energy at this instant is given by

\; \; \frac{1}{2} m \left( \; \; \frac{dy}{dt} \right)_{max}^{2} = \; \; \frac{1}{2} m \omega^{2} a^{2} = 8 \times 10^{-3} \text{joule}

or

\; \; \frac{1}{2} \times (0.1) \omega^{2} \times (0.1)^{2} = 8 \times 10^{-3}

Solving we get

\omega = \pm 4

Substituting the values of a, Z and I in the equation of

S.H.M., we get

y = 0.1 \sin\left(\pm 4t + \frac{\pi}{4}\right)

metre.

© examsnet.com

Go to Question:

Prev Question

Next Question