Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The maximum velocity of a simple harmonic motion represented by $y = 3 \sin \left( 100t + \frac{\pi}{6} \right)$ is given by
$(a) 300 (b) \frac{3 \pi}{6} (c) 100 (d) \frac{\pi}{6}$
Text Solution
Verified by ExpertsThe correct answer is:
A
$\mathbf{v}_{\max} = a \omega = 3 \times 100 = 300$
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
A body is executing simple harmonic motion with an angular frequency 2rad/s. The velocity of the bo…
A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is…
A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of t…
A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm. Its maximum speed in cm…
A particle is executing S.H.M. If its amplitude is 2 m and periodic time 2 seconds, then the maximu…
A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be
$(a) \frac{4a}{T} \quad …