Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A body of mass $1\,\mathrm{kg}$ is executing simple harmonic motion. Its displacement $y(\mathrm{cm})$ at t seconds is given by $y = 6 \sin(100t + \pi/4)$ . Its maximum kinetic energy is
Text Solution
Verified by ExpertsThe correct answer is:
B
So $a = 6cm, \omega = 100 \mathrm{rad}/\mathrm{sec}$
$K_{\max} = \frac{1}{2} m \omega^{2} a^{2} = \frac{1}{2} \times 1 \times (100)^{2} \times (6 \times 10^{-2})^{2} = 18 \, \mathrm{J}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Which of the following is a necessary and sufficient condition for S.H.M.
A particle executes simple harmonic motion along a straight line with an amplitude A. The potential…
A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement…
For a particle executing simple harmonic motion, the kinetic energy K is given by . The maximum va…
The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when…
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equa…