The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U = 5x (x – 4) J, where x is in meters. It can be concluded that
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(b, c, d)
F = 
= –[10x – 20] = 20 – 10x
acceleration a = 
= 100(2–x) = -ꞷ 2 (x – 2)
a = 0 at x = 2
So V is maximum at x = 2
This is equation of S.H.M so particle executes S.H.M
also ꞷ 2 = 100 ⇒ ꞷ = 10
T = 
= π /5 sec.
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
Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pa…
The displacement of a particle varies with time as (in cm). If its motion is S.H.M., then its maxi…
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy…
A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing …
A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept …
The amplitude of vibration of a particle is given by where and c are positive. The condition for …