Test Index

BITSAT 2018 Paper

Section: Physics

© examsnet.com

Question : 9 of 150

Marks: +1, -0

A particle of mass m executes simple harmonicmotion with amplitude a and frequency v. Theaverage kinetic energy during its motion from theposition of equilibrium to the end is

$2 \pi^2 m a^2 v^2$
$\pi^2 m a^2 v^2$
$\frac{1}{4} m a^2 v^2$
$4 \pi^2 m a^2 v^2$

Solution:

The kinetic energy of a particle executing S.H.M. is given by

K =

\frac{1}{2} m a^2 \omega^2 \sin^2 \omega t

Now, average K.E. = < K > = <

\frac{1}{2} m \omega^2 a^2 \sin^2 \omega t

>

=

\frac{1}{2} m \omega^2 a^2

<

\sin^2

ωt >

=

\frac{1}{2} m \omega^2 a^2 \left( \frac{1}{2} \right)

(Since <

\sin^2

θ > =

\frac{1}{2}

)

=

\frac{1}{4} m \omega^2 a^2

=

\frac{1}{4} m a^2 (2 \pi v)^2

(Since ω 2πv)

or < K > =

\pi^2 m a^2 v^2

© examsnet.com

Go to Question:

Prev Question

Next Question